Explanation: 1: Multiply the first equation by 2 to get $x + 4y = 28$. 2: Multiply the second equation by 6 to get $18x + y = 78$. 3: From the second new equation, $y = 78 - 18x$. 4: Substitute this into the first new equation: $x + 4(78 - 18x) = 28$. 5: $x + 312 - 72x = 28 \implies -71x = -284 \implies x=4$. 6: Substitute x=4 to find y: $y = 78 - 18(4) = 78 - 72 = 6$.
Explanation: 1: Simplify the first equation: $3x+15+7y-14=-5 \implies 3x+7y=-6$. 2: Simplify the second equation: $2x+12 - \frac{4-y}{5}=5 \implies 2x+7 = \frac{4-y}{5} \implies 10x+35 = 4-y \implies 10x+y=-31$. 3: None of the given options satisfy both simplified equations For example, testing x=-2, y=1 in the first equation gives $3(-2)+7(1) = 1$, which is not -6.
Explanation: 1: Simplify the first equation: $x + 8y = 28$. 2: Simplify the second equation: $19 = 4(x + y/4) \implies 19 = 4x + y$. 3: From the second simplified equation, $y = 19 - 4x$. 4: Substitute this into the first: $x + 8(19 - 4x) = 28 \implies x + 152 - 32x = 28 \implies -31x = -124 \implies x=4$. 5: Find y: $y = 19 - 4(4) = 3$.
Q4: The number of pencils with P is $5/3$ times the number of pencils with Q. If P has 18 pencils more than Q, then find the total number of pencils with them.
Correct Option:
Correct Answer: 72
Explanation: Let the number of pencils be P and Q 1 $P = \frac{5}{3}Q$
2 $P = Q + 18$
Substitute P: $\frac{5}{3}Q = Q + 18 \implies \frac{2}{3}Q = 18 \implies Q=27$.
Then, $P = 27 + 18 = 45$.
Total pencils = $P+Q = 45+27 = 72$.
Q5: A question paper consists of 50 questions. Each correct answer fetches three marks and one mark is deducted for each wrong answer. A student who attempted all the questions scored 90 marks. Find the number of questions answered correctly by him.
Correct Option:
Correct Answer: 35
Explanation: Let c be the number of correct answers and w be the number of wrong answers 1 $c + w = 50$
2 $3c - w = 90$
Add the two equations: $4c = 140 \implies c = 35$.
Q6: The sum of the ages of two friends A and B 18 years ago was half of the sum of their ages today. Presently, A is twice as old as B. What is the present age of A (in years)?
Correct Option:
Correct Answer: 48
Explanation: Let the present ages be A and B 1 $(A-18) + (B-18) = \frac{1}{2}(A+B) \implies A+B-36 = \frac{A+B}{2} \implies 2A+2B-72=A+B \implies A+B=72$ 2 $A=2B$ Substitute A: $2B+B = 72 \implies 3B = 72 \implies B=24$.
Present age of A is $2 \times 24 = 48 years.
Q7: A student was asked to find $3/7^{th}$ of a number and he instead multiplied it by $7/3$. As a result, he got an answer, which was more than the correct answer by 1680. What was the number?
882
273
840
1684
Correct Option: A
Correct Answer: 882
Explanation: Let the number be N The difference is $(\frac{7}{3}N) - (\frac{3}{7}N) = 1680$.
$(\frac{49-9}{21})N = 1680 \implies \frac{40}{21}N = 1680$.
$N = 1680 \times \frac{21}{40} = 42 \times 21 = 882$.
Q8: A boy has a total of ₹14 in denominations of 25 paise and 20 paise coins. If the numbers of coins of the two denominations were swapped, the total value of coins would be ₹1 less. Find the total number of coins.
Correct Option:
Correct Answer: 60
Explanation: Let q be the number of 25p coins and t be the number of 20p coins Value is in paise 1 $25q + 20t = 1400 \implies 5q + 4t = 280$
2 $20q + 25t = 1300 \implies 4q + 5t = 260$
Add the two simplified equations: $9q + 9t = 540$.
Divide by 9: $q + t = 60$ The total number of coins is 60.
Q9: Govind is four times as old as Ganesh is. 20 years hence, Govind's age will be twice that of Ganesh's age. Find Ganesh's present age. (in years)
20
10
15
30
Correct Option: B
Correct Answer: 10
Explanation: Let Govind's age be Go and Ganesh's be Ga 1 $Go = 4Ga$
2 $Go+20 = 2(Ga+20)$
Substitute Go: $4Ga+20 = 2Ga+40 \implies 2Ga = 20 \implies Ga=10$.
Q10: Five years ago, Alok's age was five times Bharan's age. Five years hence, Alok's age will be thrice that of Bharan's age. Find Bharan's present age. (in years)
Correct Option:
Correct Answer: 15
Explanation: Let present ages be A and B 1 $A-5 = 5(B-5) \implies A-5B = -20$
2 $A+5 = 3(B+5) \implies A-3B = 10$
Subtract the first equation from the second: $2B = 30 \implies B=15$.
Q11: Praveen's present age is twice that of Mahesh's age four years ago. Eight years hence, Praveen would be twice as old as Mahesh is today. Find the sum of their present ages.
36 years
44 years
64 years
Cannot be determined
Correct Option: D
Correct Answer: Cannot be determined
Explanation: Let present ages be P and M 1 $P = 2(M-4) \implies P = 2M-8$
2 $P+8 = 2M$
Substituting P from the first equation into the second gives $(2M-8)+8 = 2M \implies 2M = 2M$ This is an identity, meaning the equations are dependent and there is no unique solution.
Q12: Alok's age is $5/3$ times of Alakhnanda's age. Alakhnanda is now 3 times as old as she was, when Alok was as old as Alakhnanda is today. Find Alok's age when Alakhnanda was half as old as Alok is now.
60
50
40
Data insufficient
Correct Option: D
Correct Answer: Data insufficient
Explanation: Let present ages be A and N 1 $A = \frac{5}{3}N \implies 3A = 5N$
2 The time when Alok was N years old was A-N years ago At that time, Alakhnanda's age was N-(A-N) = 2N-A The condition is $N = 3(2N-A) \implies N = 6N-3A \implies 3A=5N$ Both conditions give the same relationship, so individual ages cannot be determined.
Q13: The sum of the ages of Ajay and Bala, 20 years ago was five-ninth the sum of their present ages. Ajay's present age exceeds that of Bala by 20 years. Find the present age of Ajay. (in years)
Correct Option:
Correct Answer: 55
Explanation: Let present ages be J and B Let the present sum be S = J+B 1 (J-20)+(B-20) = \frac{5}{9}(J+B) \implies S-40 = \frac{5}{9}S \implies 9S-360=5S \implies 4S=360 \implies S=90$ So $J+B=90$ 2 $J = B+20$ Substitute J: $(B+20)+B = 90 \implies 2B = 70 \implies B=35$ Ajay's age is $J=35+20=55$.
Q14: Ten years ago, the age of a man was 20 years less than 6 times his son's age. Ten years hence, his age will be 30 years less than thrice his son's age. After how many years from now will their combined age be 90 years?
5
10
15
20
Correct Option: B
Correct Answer: 10
Explanation: Let present ages be M and S 1 $M-10 = 6(S-10)-20 \implies M-6S = -70$
2 $M+10 = 3(S+10)-30 \implies M-3S = -10$
Solving gives S=20 and M=50 Their current combined age is 70 To reach a combined age of 90, they need to age a combined 20 years Since they both age, it will take $20/2 = 10$ years.
Q15: A two-digit number is formed by either subtracting 16 from eight times the sum of the digits or by adding 20 to 22 times the difference of the digits. Find the number.
24
48
64
82
Correct Option: C
Correct Answer: 64
Explanation: Let the number be 10t+u 1 $10t+u = 8(t+u)-16 \implies 2t-7u = -16$
2 $10t+u = 22(t-u)+20 \implies -12t+23u = 20$
Multiply the first equation by 6: $12t-42u = -96$ Add this to the second equation: $-19u = -76 \implies u=4$.
Substitute u=4 into the first equation: $2t - 7(4) = -16 \implies 2t=12 \implies t=6$ The number is 64.
Q16: If $x/4$ years ago, Alok was 14 years old and $x/4$ years from now he will be 4x years old, how old will he be 5x years from now? (in years)
Correct Option:
Correct Answer: 35
Explanation: Let Alok's present age be A 1 $A - x/4 = 14$
2 $A + x/4 = 4x$
Adding both equations gives $2A = 14 + 4x \implies A = 7 + 2x$.
Substituting A in the first equation: $(7+2x) - x/4 = 14 \implies 7x/4 = 7 \implies x=4$.
Alok's present age $A = 7 + 2(4) = 15$.
His age 5x (i.e., 20) years from now will be $15 + 20 = 35$.
Q17: Two boys and two girls went to a movie. They found that there were only two tickets available in the counter and they bought them. For purchasing the remaining two tickets (in black), they spent ₹50 more for each ticket than the actual price. At the end they found that each person had spent ₹60 for the ticket as his/her share. Find the actual price (in ₹) of each ticket.
Correct Option:
Correct Answer: 35
Explanation: There are 4 people, and the average cost per person was ₹60 Total amount spent = $4 \times 60 = ₹240$.
Let the actual price of a ticket be P Total cost = (2 \times P) + 2 \times (P + 50).
$240 = 2P + 2P + 100 \implies 140 = 4P \implies P = 35$.
Q18: There are two two-digit numbers such that the tens digit of the first number is $3/2$ times the tens digit of the second number, while the sum of the two numbers is 158. Which of the following can be the difference between them?
58
71
36
40
Correct Option: C
Correct Answer: 36
Explanation: Let the numbers be $N_1 = 10a+b$ and $N_2 = 10c+d$ $a = \frac{3}{2}c$ The only digit pair (a,c) that allows for a sum of 158 is (9,6).
$N_1+N_2 = (90+b) + (60+d) = 158 \implies b+d=8$ The difference is $N_1-N_2 = (90+b) - (60+d) = 30 + (b-d)$ If we choose b=7 and d=1 (which satisfies b+d=8), the difference is $30 + (7-1) = 36$.
Q19: In a three-digit number, the difference between hundreds digit and the tens digit is equal to the difference between the tens digit and the units digit. If the sum of the digits is 9, how many numbers satisfy the given condition?
Correct Option:
Correct Answer: 6
Explanation: Let the digits be h, t, u 1 h-t = t-u \implies h+u=2t (The digits are in an arithmetic progression).
2 h+t+u = 9.
Substitute h+u from (1) into (2): $(2t) + t = 9 \implies 3t=9 \implies t=3$.
Now, $h+u=6$ The possible numbers (where h≠0) are: 135, 234, 333, 432, 531, 630 There are 6 such numbers.
Q20: The difference between a three-digit number and the number formed by reversing its digits is 792. The sum of its digits is 18 and the hundreds digit is 9 times its units digit. Find the number.
Correct Option:
Correct Answer: 981
Explanation: Let the number be 100h+10t+u.
1 $99(h-u) = 792 \implies h-u = 8$.
2 h=9u.
Solving these two gives u=1 and h=9.
3 h+t+u = 18 \implies 9+t+1 = 18 \implies t=8.
The number is 981.
Q21: Mr. Ram distributed a total of 225 chocolates among his sons A, B, C and D. The number of chocolates he gave to A and D together is twice the number of chocolates he gave to B and C together. If B received 15 more chocolates than C, find the number of chocolates C received.
Correct Option:
Correct Answer: 30
Explanation: Total: (A+D) + (B+C) = 225.
1 A+D = 2(B+C) Substitute this into the total: $2(B+C) + (B+C) = 225 \implies 3(B+C) = 225 \implies B+C = 75$.
2 B = C+15.
Substitute B: $(C+15)+C=75 \implies 2C=60 \implies C=30$.
Q22: If the numerator of a fraction is increased by two and the denominator by one, the fraction becomes $13/15$. If the numerator and the denominator are each decreased by four, the fraction becomes 4/5. Find the fraction.
$9/24$
$13/19$
$24/29$
None of these
Correct Option: C
Correct Answer: 24/29
Explanation: Let the fraction be N/D.
1 (N+2)/(D+1) = 13/15 \implies 15N - 13D = -17.
2 (N-4)/(D-4) = 4/5 \implies 5N - 4D = 4.
Multiplying the second equation by 3 gives 15N - 12D = 12.
Subtracting this from the first gives -D = -29 \implies D=29.
Then 5N - 4(29) = 4 \implies 5N = 120 \implies N=24 The fraction is 24/29.
Q23: Ajay and Sita are two of Mr.Sharma's children. Ajay has half as many brothers as sisters. Sita has as many brothers as sisters. Find the number of children Mr.Sharma has.
Correct Option:
Correct Answer: 7
Explanation: Let b be the number of boys and g be the number of girls.
1 From Ajay's (a boy) perspective: (b-1) = \frac{1}{2}g \implies 2b-2 = g.
2 From Sita's (a girl) perspective: b = g-1 \implies g=b+1.
Equating the expressions for g: $2b-2 = b+1 \implies b=3$.
Then $g = 3+1 = 4$ Total children = 3+4=7.
Q24: Rohan went to the market to buy 10 kg of each of oranges, mangoes, bananas and grapes. The cost of 5 kg oranges and 2 kg mangoes together was ₹310. The cost of 3 kg mangoes and 3.5 kg bananas together was ₹230. The cost of 1.5 kg bananas and 5 kg grapes together was ₹160. Find the total amount spent by Rohan (in ₹).
Correct Option:
Correct Answer: 1400
Explanation: Let the costs per kg be O, M, B, G We have:
1 5O + 2M = 310
2 3M + 3.5B = 230
3 1.5B + 5G = 160
Adding all three equations: 5O + (2M+3M) + (3.5B+1.5B) + 5G = 310+230+160.
$5O + 5M + 5B + 5G = 700 \implies O+M+B+G = 140$.
Total cost for 10kg of each is $10(O+M+B+G) = 10(140) = 1400$.
Q25: If he bought less than 15 bananas, how many oranges did he buy?
Correct Option:
Correct Answer: 18
Explanation: Given: o > b > a, a, b, o >= 13, a+b+o = 45.
Condition: b < 15 Since b >= 13, b must be 13 or 14 If b=13, then a must be <13, which violates a>=13.
Therefore, b must be 14 Then a must be 13 (as a < b and a >= 13).
$o = 45 - a - b = 45 - 13 - 14 = 18$.
Q26: The cost of an apple, a banana and an orange is ₹5, ₹4 and ₹3 respectively. What is the minimum possible expenditure (in ₹) that Arjun could have incurred?
Correct Option:
Correct Answer: 175
Explanation: From the conditions o > b > a, a,b,o >= 13, a+b+o = 45, we want to minimize 5a + 4b + 3o.
To minimize the cost, we should buy the fewest expensive items (apples) and the most cheap items (oranges).
This means we should choose the smallest possible values for a and b.
Min a=13, min b=14 This gives o=18.
Expenditure = $5(13) + 4(14) + 3(18) = 65 + 56 + 54 = 175$.
Q27: If a, b, c and d satisfy the equations $a+7b+3c+5d=0$, $8a+4b+6c+2d=-16$, $a+6b+4c+8d=16$ and $5a+3b+7c+d=-16$ then (a+d)(b+c) equals
0
16
-16
-64
Correct Option: C
Correct Answer: -16
Explanation: Let the four equations be (1), (2), (3), (4).
Add (1) and (4): $6a+10b+10c+6d = -16 \implies 3(a+d)+5(b+c) = -8$ (Eq A)
Add (2) and (3): $9a+10b+10c+10d = 0 \implies 9a+10(b+c)+10d=0$ (Eq B)
Let $X=a+d$ and $Y=b+c$ Then $3X+5Y=-8$ A possible integer solution is X=4, Y=-4 Test this in Eq B: $9a+10(-4)+10d=0 \implies 9a+10d=40$ Using $a+d=4 \implies d=4-a$, we get $9a+10(4-a)=40 \implies -a=0 \implies a=0$ This is consistent So $a+d=4$ and $b+c=-4$ The product is $(4)(-4) = -16$.
Q28: Considering the equations $2x-3y=8$ and $px-qy=66$ answer the following questions: (i) Find 4(p+q) if the equations above have infinite solutions. (ii) Find p if q=9 and the equations above have no solution.
Correct Option:
Correct Answer: (i) 165
(ii) 6
Explanation: For a system $a_1x+b_1y=c_1, a_2x+b_2y=c_2$:
(i) Infinite Solutions: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ $\frac{2}{p} = \frac{-3}{-q} = \frac{8}{66} \implies \frac{2}{p} = \frac{3}{q} = \frac{4}{33}$ This gives $p = 33/2$ and $q = 99/4$ $4(p+q) = 4(\frac{66}{4}+\frac{99}{4}) = 165$ (ii) No Solution: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ $\frac{2}{p} = \frac{-3}{-9} \implies \frac{2}{p} = \frac{1}{3} \implies p = 6$ (Check: $\frac{1}{3} \neq \frac{8}{66}$, condition holds).
Q29: nan
Correct Option:
Correct Answer:
Explanation:
Q30: In a four-digit number with distinct digits, the sum of the middle digits equals the sum of the extreme digits. The sum of its second and fourth digits equals five times the sum of its other two digits. If the sum of its digits is 18, what is the sum of all the possible values of the hundreds digit?
21
24
27
30
Correct Option: B
Correct Answer: 24
Explanation: Let the number be abcd.
1 $b+c = a+d$
2 $a+b+c+d=18 \implies 2(b+c)=18 \implies b+c=9$ This also means $a+d=9$.
3 $b+d = 5(a+c)
Substitute c=9-b and d=9-a into the third equation: $b+(9-a) = 5(a+(9-b)) \implies 6b-6a=36 \implies b-a=6$.
Possible pairs for (a,b) are (1,7), (2,8), (3,9) These give the numbers 1728, 2817, and 3906.
The possible values for the hundreds digit (b) are 7, 8, and 9 The sum is $7+8+9=24$.
Q31: A two-digit number is obtained by either subtracting 12 from four times the sum of its digits or by adding 6 to twice the difference of its digits. Find the number.
16
28
39
Cannot be determined
Correct Option: A
Correct Answer: 16
Explanation: Let the number be 10t+u.
1 $10t+u = 4(t+u)-12 \implies 6t-3u=-12 \implies 2t-u=-4$.
2 $10t+u = 2(u-t)+6 \implies 12t-u=6$ (Assuming u>t).
Subtracting the first equation from the second gives $10t=10 \implies t=1$.
From 2t-u=-4, we get $2(1)-u=-4 \implies u=6$ The number is 16.
Q32: Ramu has some chocolate boxes with him to sell. He sells them either as full boxes or half boxes. The first customer buys half a box more than half the number of boxes with Ramu. The second customer buys half a box more than half the remaining number of boxes with him. Ramu continues to sell in this manner to eight other customers. He is left with no boxes to sell after that. How many chocolate boxes did Ramu have in the beginning?
511
513
1023
1025
Correct Option: C
Correct Answer: 1023
Explanation: This problem involves 10 customers in total Let C_n be the number of boxes before customer n The number left after customer n is C_n - (0.5 C_n + 0.5) = 0.5 C_n - 0.5.
So, C_{n+1} = 0.5 C_n - 0.5, which means $C_n = 2 C_{n+1} + 1$.
Working backwards from C_{11}=0 (after 10 customers):
$C_{10} = 2(0)+1=1$.
$C_9 = 2(1)+1=3$.
$C_8 = 2(3)+1=7$.
The pattern is $C_{11-k} = 2^k - 1$ For the beginning amount (C_1), k=10.
Initial boxes = $2^{10} - 1 = 1024 - 1 = 1023$.
Q33: Alok went to a casino to play a card game. He played 10 rounds of that game. In each round, he doubled his amount and then gave x to his friend. After 10 rounds, he had ₹1023. Find the sum of the digits of the least possible value of x. (All the amounts involved (in rupees) are integers)
Correct Option:
Correct Answer: 6
Explanation: Let the initial amount be $A_0$ The amount after n rounds is $A_n = 2^n A_0 - (2^n - 1)x$.
After 10 rounds, $A_{10} = 2^{10}A_0 - (2^{10}-1)x = 1024A_0 - 1023x$.
We are given $1024A_0 - 1023x = 1023$.
$1024A_0 = 1023(x+1)$.
Since 1024 and 1023 are coprime, $x+1$ must be a multiple of 1024 The least positive integer value for x occurs when $x+1=1024 \implies x=1023$.
The sum of the digits of x is $1+0+2+3=6$.
Q34: How many possibilities exist for the actual number of toys sold?
Correct Option:
Correct Answer: 1
Explanation: Let the actual number of toys be N=10a+b The mistaken number is N'=10b+a The computer stock showed 81 more items left, which means the recorded number of sales was 81 less than actual.
$N - N' = 81$
$(10a+b) - (10b+a) = 81$
$9a - 9b = 81 \implies a - b = 9$.
Since a and b are single digits and a cannot be 0, the only possibility is a=9, b=0 This gives only 1 possible value for the number of toys sold (N=90).
Q35: If the faulty calculations show a total sale of ₹486, what was the actual selling price (in ₹) of each toy?
Correct Option:
Correct Answer: 45
Explanation: From the previous analysis, the actual number of toys sold was N=90, so the faulty number sold was N'=09=9.
Faulty Sale = Faulty Number × Faulty Price = 486.
$9 \times P' = 486 \implies P' = 54$.
The faulty price P' = 10d+c = 54, so d=5 and c=4.
The actual selling price is P = 10c+d = 10(4)+5 = 45.
Q36: Raja went to a casino to play a card game. He played 3 rounds of the game. In each round he doubled the amount he had with him and gave X to his friend at the end of the round. The amount he had with him at the end of the third round after giving X to his friend was ₹140 more than the sum of the amounts with him at the end of the previous rounds after giving X to his friend. The amount with him at the end of the second round after giving X to his friend was ₹160 more than the amount he had with him at the end of the first round after giving X to his friend. Find the value of X.
10
20
30
40
Correct Option: B
Correct Answer: 20
Explanation: Let A1, A2, A3 be the amounts after rounds 1, 2, and 3.
1 $A2 = 2A1 - X$
2 $A3 = 2A2 - X$
Given conditions:
3 $A2 = A1 + 160$
4 $A3 = (A1+A2)+140$
From (1) and (3): $A1+160 = 2A1-X \implies A1 = X+160$.
Now find A2 and A3 in terms of X: $A2 = (X+160)+160 = X+320$.
$A3 = 2(X+320)-X = X+640$.
Substitute into (4): $X+640 = (X+160)+(X+320)+140 \implies X+640 = 2X+620 \implies X=20$.
Q37: If $x+2y+3z=14$, then find the value of z.
I. $x+3y+z=14$
II. $3x+y+2z=11$
Correct Option: C
Correct Answer: (C)
Explanation: With the initial equation, we have 1 equation and 3 unknowns We need two more independent equations.
Statement I alone gives a second equation Subtracting it from the given equation yields y-2z=0, but we cannot find z Insufficient.
Statement II alone gives a second equation We still have 2 equations and 3 unknowns Insufficient.
Using both statements I and II together with the initial equation gives a system of 3 linear equations with 3 variables, which can be solved for a unique value of z Sufficient.
Q38: Guru had some one-rupee coins, 50-rupee notes and 100-rupee notes. He exchanged all his coins for 50-rupee and 100-rupee notes (not by value, only by number). After the exchange, Guru has ₹500. How many 50-rupee notes does he have after the exchange?
I. He has not more than 6 notes after the exchange.
II. He has not less than 6 notes after the exchange.
Correct Option: A
Correct Answer: (A)
Explanation: Let f be the number of ₹50 notes and h be the number of ₹100 notes after exchange The total value is $50f + 100h = 500 \implies f+2h=10$ Possible integer solutions (f, h) are (8,1), (6,2), (4,3), (2,4).
Statement I: Total notes (f+h) <= 6 This is only satisfied by the solution (2,4), where f+h=6 So f=2 Sufficient.
Statement II: Total notes (f+h) >= 6 This is satisfied by all four solutions So f could be 2, 4, 6, or 8 Insufficient.
Q39: What is my age?
I. Five years ago, my sister's age was half of my age.
II. Five years from now, my sister's age will be three fourths of my age.
Correct Option: C
Correct Answer: (C)
Explanation: Let my age be M and my sister's be S.
Statement I: S-5 = (M-5)/2 One equation, two variables Insufficient.
Statement II: S+5 = 3(M+5)/4 One equation, two variables Insufficient.
Together, we have a system of two independent linear equations with two variables (2S-M=5 and 4S-3M=-5), which can be solved for unique values of M and S Sufficient.
Q40: How many questions did I attempt in a maths test having 25 questions?
I. I scored 16 marks.
II. For every correct answer I got 1 mark while for every incorrect answer I lost $\frac{1}{4}$ mark.
Correct Option: D
Correct Answer: (D)
Explanation: Let c, w, u be correct, wrong, unanswered We need c+w c+w+u=25.
Statement I gives the score but not the marking scheme Insufficient.
Statement II gives the marking scheme but not the score Insufficient.
Together: c - (w/4) = 16 \implies 4c-w=64 We also have c+w \leq 25.
If c=16, w=0, so c+w=16 If c=17, w=4, so c+w=21 Since there are multiple possible values for the number of attempted questions, the information is insufficient.
Q41: If $3x+7y=19$, then find the value of y.
I. $6x+14y=38$
II. $9x-20y=16$
Correct Option: A
Correct Answer: (A)
Explanation: Statement I: $6x+14y=38$ simplifies to $3x+7y=19$ This is the same as the given equation and provides no new information Insufficient.
Statement II: We now have a system of two different linear equations: 3x+7y=19 and 9x-20y=16 This system can be solved for a unique value of y Sufficient.
Since the question can be answered by II alone but not by I alone, the correct option is (A).
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