Explanation: 1: Let <span class="math-inline">a \= \\frac\{1\}\{2x\+3y\}</span> and <span class="math-inline">b \= \\frac\{1\}\{3x\+2y\}</span> The equations become:
(i) <span class="math-inline">16a \- 7b \= 1</span>
(ii) <span class="math-inline">\\frac\{8\}\{3\}a \+ 21b \= \\frac\{10\}\{3\}</span> which simplifies to <span class="math-inline">8a \+ 63b \= 10</span>. 2: Multiply the simplified (ii) by 2 to get <span class="math-inline">16a \+ 126b \= 20</span>. 3: Subtract (i) from this: <span class="math-inline">\(16a\+126b\) \- \(16a\-7b\) \= 20\-1</span>, which gives <span class="math-inline">133b \= 19</span>, so <span class="math-inline">b \= \\frac\{1\}\{7\}</span>. 4: Substitute b in (i): <span class="math-inline">16a \- 7\(\\frac\{1\}\{7\}\) \= 1</span>, which gives <span class="math-inline">16a\=2</span>, so <span class="math-inline">a\=\\frac\{1\}\{8\}</span>. 5: Substitute back: <span class="math-inline">2x\+3y\=8</span> and <span class="math-inline">3x\+2y\=7</span> Solving this system gives <span class="math-inline">x\=1, y\=2</span>.
Q3: A rope of 77 meters is cut into two pieces such that the length of one piece is <span class="math-inline">4/7^\{th\}</span> of the other. What is the length of <span class="math-inline">3/14^\{th\}</span> of the longer piece? (in m)
Correct Option:
Correct Answer: 10.5
Explanation: 1: Let the pieces be x and y <span class="math-inline">x\+y\=77</span> and <span class="math-inline">x \= \\frac\{4\}\{7\}y</span>. 2: Substitute x in the first equation: <span class="math-inline">\\frac\{4\}\{7\}y \+ y \= 77 \\implies \\frac\{11\}\{7\}y \= 77 \\implies y \= 49</span>. 3: The other piece is <span class="math-inline">x \= 77 \- 49 \= 28</span> The longer piece is 49 m. 4: Calculate <span class="math-inline">3/14^\{th\}</span> of the longer piece: $\frac{3}{14} \times 49 = \frac{3 \times 7}{2} = 10.5 m.
Q4: Anand has only 10 paise and 25 paise coins with him. If he has 70 coins in all worth ₹10 with him, how many 25 paise coins does he have?
20
25
40
50
Correct Option: A
Correct Answer: 20
Explanation: 1: Let t be the number of 10p coins and q be the number of 25p coins <span class="math-inline">t\+q\=70</span>. 2: The total value is <span class="math-inline">10t \+ 25q \= 1000</span> (since ₹10 = 1000 paise). 3: From the first equation, <span class="math-inline">t\=70\-q</span> Substitute this into the value equation: <span class="math-inline">10\(70\-q\) \+ 25q \= 1000</span>. 4: Simplify and solve for q: <span class="math-inline">700 \- 10q \+ 25q \= 1000 \\implies 15q \= 300 \\implies q \= 20</span> He has 20 coins of 25 paise.
Q5: Find the greater of the two numbers such that their sum is 200 and the difference of their squares is 8000.
80
100
120
140
Correct Option: C
Correct Answer: 120
Explanation: 1: Let the numbers be x and y <span class="math-inline">x\+y\=200</span>. 2: The difference of their squares is <span class="math-inline">x^2 \- y^2 \= 8000</span>. 3: Factor the difference of squares: <span class="math-inline">\(x\-y\)\(x\+y\) \= 8000</span>. 4: Substitute the sum: <span class="math-inline">\(x\-y\)\(200\) \= 8000 \\implies x\-y \= 40</span>. 5: Solve the system <span class="math-inline">x\+y\=200</span> and <span class="math-inline">x\-y\=40</span> Adding them gives <span class="math-inline">2x\=240 \\implies x\=120</span> The other number is <span class="math-inline">y\=80</span> The greater number is 120.
Q6: A fraction becomes <span class="math-inline">1/2</span>, if its numerator is increased by 1 and the denominator by 3. It becomes <span class="math-inline">2/5</span> if the numerator is increased by 2 and the denominator by 7. Find the fraction.
Explanation: 1: Let the fraction be x/y From the first condition, <span class="math-inline">\\frac\{x\+1\}\{y\+3\} \= \\frac\{1\}\{2\} \\implies 2x\+2 \= y\+3 \\implies 2x\-y\=1</span>. 2: From the second condition, <span class="math-inline">\\frac\{x\+2\}\{y\+7\} \= \\frac\{2\}\{5\} \\implies 5x\+10 \= 2y\+14 \\implies 5x\-2y\=4</span>. 3: From <span class="math-inline">2x\-y\=1</span>, we get <span class="math-inline">y\=2x\-1</span> Substitute into the second equation: <span class="math-inline">5x\-2\(2x\-1\)\=4</span>. 4: Solve for x: <span class="math-inline">5x\-4x\+2\=4 \\implies x\=2</span> Then <span class="math-inline">y\=2\(2\)\-1\=3</span> The fraction is <span class="math-inline">\\frac\{2\}\{3\}</span>.
Q7: There is some money with Ajay and some with Vijay. If Ajay gives ₹30 to Vijay, then the amounts with them would be interchanged. Instead, if Vijay gives ₹20 to Ajay, then Ajay would have ₹70 more than Vijay would have. Find the amount that Ajay has.
40
50
70
Cannot be determined
Correct Option: D
Correct Answer:
Explanation:
Q8: How many two-digit numbers with their tens digit greater than their units digit, have the sum of their digits equal to twice their difference?
Correct Option:
Correct Answer: 3
Explanation: 1: Let the number be 10t+u, with t>u. 2: The condition is <span class="math-inline">t\+u \= 2\(t\-u\)</span>. 3: Simplify the equation: <span class="math-inline">t\+u \= 2t\-2u \\implies 3u\=t</span>. 4: We need to find pairs of digits (t, u) where t=3u and t>u.
If <span class="math-inline">u\=1, t\=3</span> Number: 31 (3>1 OK).
If <span class="math-inline">u\=2, t\=6</span> Number: 62 (6>2 OK).
If <span class="math-inline">u\=3, t\=9</span> Number: 93 (9>3 OK).
If <span class="math-inline">u\=4, t\=12</span> (Not a digit).
There are 3 such numbers.
Q9: A two-digit number is such that the sum of its digits is thrice the difference of its digits. If the number exceeds the number formed by reversing its digits by 36, find the number.
Correct Option:
Correct Answer: 84
Explanation: 1: Let the number be 10t+u Reversal is 10u+t. 2: From the second condition, <span class="math-inline">\(10t\+u\)\-\(10u\+t\)\=36 \\implies 9t\-9u\=36 \\implies t\-u\=4</span> This means t>u. 3: From the first condition, <span class="math-inline">t\+u \= 3\(t\-u\)</span>. 4: Substitute <span class="math-inline">t\-u\=4</span> into the first condition: <span class="math-inline">t\+u \= 3\(4\) \\implies t\+u\=12</span>. 5: Solve the system <span class="math-inline">t\-u\=4</span> and <span class="math-inline">t\+u\=12</span> Adding them gives <span class="math-inline">2t\=16 \\implies t\=8</span> Then <span class="math-inline">u\=4</span> The number is 84.
Q10: The difference between a three-digit number and the number formed by reversing its digits is 297. The sum of the units and the tens digits is the same as the difference of the hundreds and the units digits. Also, the hundreds digit is twice the units digit. Find the number.
242
342
603
884
Correct Option: C
Correct Answer: 603
Explanation: 1: Let the number be 100h+10t+u The difference with its reverse is <span class="math-inline">99\(h\-u\) \= 297 \\implies h\-u\=3</span>. 2: The hundreds digit is twice the units digit: <span class="math-inline">h\=2u</span>. 3: Substitute h=2u into the first result: <span class="math-inline">2u\-u\=3 \\implies u\=3</span> Then <span class="math-inline">h\=2\(3\)\=6</span>. 4: The sum of units and tens equals the difference of hundreds and units: <span class="math-inline">u\+t \= h\-u</span>. 5: Substitute known values: <span class="math-inline">3\+t \= 6\-3 \\implies 3\+t\=3 \\implies t\=0</span> The number is <span class="math-inline">100\(6\)\+10\(0\)\+3 \= 603</span>.
Q11: Four years ago, a man was thrice as old as his son. Eight years hence, the man will be twice as old as his son. What is the present age (in years) of the son?
Correct Option:
Correct Answer: 16
Explanation: Let the present ages of the man and son be M and S From the given conditions:
1 $M - 4 = 3(S - 4) \implies M - 3S = -8$
2 $M + 8 = 2(S + 8) \implies M - 2S = 8$
Subtracting the first equation from the second gives S = 16.
Q12: The present average age of Ram and his wife Sita and their daughter is 35 years. Fifteen years from now, the age of Sita will be equal to the sum of the present ages of Ram and the daughter. Find the present age (in years) of Sita.
Correct Option:
Correct Answer: 45
Explanation: Let the present ages be R, S, and D 1 $(R + S + D)/3 = 35 \implies R + S + D = 105$
2 $S + 15 = R + D
Substitute (R + D) from the second equation into the first: $(S + 15) + S = 105 \implies 2S = 90 \implies S = 45$.
Q13: X says to Y, "I am twice as old as you were when I was as old as you are." The sum of their present ages is 63 years. Find the present age of X.
24 years
39 years
36 years
42 years
Correct Option: C
Correct Answer: 36 years
Explanation: Let the present ages of X and Y be x and y.
1 $x + y = 63$
2 The time when X was as old as Y is (x - y) years ago At that time, Y's age was $y - (x - y) = 2y - x$.
The second condition translates to $x = 2(2y - x) \implies 3x = 4y$.
Solving the two equations gives $x = 36$ and $y = 27$.
Q14: Six years ago, the age of a person was two years more than five times the age of his son. Four years hence, his age will be two years less than three times the age of his son. After how many years from now will their combined age be 100 years?
48 years
14 years
19 years
38 years
Correct Option: C
Correct Answer: 19 years
Explanation: Let the present ages be P and S.
1 $P-6 = 5(S-6)+2 \implies P-5S = -22$
2 $P+4 = 3(S+4)-2 \implies P-3S = 6$
Solving this system gives S=14 and P=48 Their current combined age is 14+48=62.
Let y be the required years $(48+y) + (14+y) = 100 \implies 62 + 2y = 100 \implies 2y=38 \implies y=19$.
Q15: (a) If the following three equations form a system of dependent equations, what is the value of p? I. $3x+2y-7z=56$ II. $5x+3y+z=16$ III. px+12y-19z=200
Correct Option:
Correct Answer: 19
Explanation: For the system to be dependent, one equation must be a linear combination of the others, i.e., Eq III = a(Eq I) + b(Eq II).
By comparing the coefficients of y and z:
• $2a + 3b = 12$
• $-7a + b = -19$
Solving this gives a=3 and b=2.
Now, compare the x-coefficients: p = 3a + 5b = 3(3) + 5(2) = 19.
Q16: Find k if the given system of equations has infinite solutions. $2x+ky=1+2y$ and $kx+12y=3$
Correct Option:
Correct Answer: 6
Explanation: Rearrange the first equation to $2x + (k-2)y = 1$.
For infinite solutions, the ratio of coefficients must be equal: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
$\frac{2}{k} = \frac{k-2}{12} = \frac{1}{3}$.
From $\frac{2}{k} = \frac{1}{3}$, we get $k=6$ This value also satisfies $\frac{k-2}{12} = \frac{4}{12} = \frac{1}{3}$.
Q17: Find the value of k if the equations $4x+5y=32$ and $12x+15y=2k$ are not inconsistent.
Correct Option:
Correct Answer: 48
Explanation: "Not inconsistent" means there is at least one solution Checking the coefficients: $\frac{4}{12} = \frac{1}{3}$ and $\frac{5}{15} = \frac{1}{3}$ Since the ratios of x and y coefficients are equal, the lines are either parallel or coincident For a solution to exist, they must be coincident The condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
$\frac{1}{3} = \frac{32}{2k} \implies 2k = 96 \implies k = 48$.
Q18: The cost of two balls, three bats and eight pairs of gloves is ₹2500, while the cost of four balls, five bats and ten pairs of gloves is ₹4000. Find the cost of each bat.
350
500
800
Cannot be determined
Correct Option: D
Correct Answer: Cannot be determined
Explanation: Let the costs be b, t, and g.
1 $2b + 3t + 8g = 2500$
2 $4b + 5t + 10g = 4000$
Multiplying the first equation by 2 gives $4b + 6t + 16g = 5000$.
Subtracting the second equation from this yields $t + 6g = 1000$.
Since there are two variables in this final equation, the individual cost of a bat (t) cannot be found.
Q19: The cost of three pens, four rulers and five refills is ₹75 while that of ten refills, six pens and seven rulers is ₹138. Find the cost of three pens, one ruler and five refills.
39
42
44
Cannot be determined
Correct Option: A
Correct Answer: 39
Explanation: Let the costs be p, r, and f.
1 $3p + 4r + 5f = 75$
2 $6p + 7r + 10f = 138$
Multiply the first equation by 2: $6p + 8r + 10f = 150$.
Subtract the second equation from this new one: $r = 12$.
We need to find $C = 3p + r + 5f$ We can rewrite the first equation as $(3p + r + 5f) + 3r = 75$.
$C + 3(12) = 75 \implies C = 39$.
Q20: The cost of two pencils, one eraser and three sharpeners is ₹23. The cost of six pencils, three erasers and one sharpener is ₹29. The cost of 14 pencils, seven erasers and seven sharpeners is ₹91. Find the cost of each pencil.
3
5
4
Cannot be determined
Correct Option: D
Correct Answer: Cannot be determined
Explanation: Let the costs be p, e, and s.
1 $2p + e + 3s = 23$
2 $6p + 3e + s = 29$
3 $14p + 7e + 7s = 91$, which simplifies to $2p + e + s = 13$.
Subtracting the simplified third equation from the first gives $2s = 10 \implies s = 5$.
Substituting $s=5$ into the simplified third equation gives $2p + e = 8$.
Substituting $s=5$ into the second equation gives $6p + 3e = 24$, which simplifies to $2p + e = 8$.
Since we get the same equation twice, we cannot determine the individual costs of a pencil and an eraser.
Q21: A bag has a total of 120 notes in denominations of ₹2, 5 and 10. The total value of the notes in the bag is ₹760. If there were twice as many ₹5 notes, the total value of the notes would be ₹960. Find the number of ₹10 notes in the bag.
Correct Option:
Correct Answer: 50
Explanation: Let the number of notes be x (₹2), y (₹5), and z (₹10).
1 $x+y+z=120$
2 $2x+5y+10z=760$
If the ₹5 notes are doubled, the value increases by $5y$ The total value increases by ₹960 - ₹760 = ₹200.
So, $5y=200 \implies y=40$.
Substituting y=40 into the first two equations gives:
$x+z=80$ and $2x+10z=560 \implies x+5z=280$.
Solving this system gives $4z=200 \implies z=50$.
Q22: Rohan went to a stationery shop to purchase pens, erasers and rulers. He purchased more number of pens than erasers and more number of erasers than rulers. He purchased at least 10 items of each type. The total number of items purchased is 35. How many rulers did Rohan purchase?
Correct Option:
Correct Answer: 10
Explanation: Let the number of pens, erasers, and rulers be p, e, and r.
We are given: p > e > r, p, e, r >= 10, and p + e + r = 35.
Since r is the smallest count and must be at least 10, let's test r=10.
If r=10, then e must be at least 11 p must be at least 12 The minimum sum is 10+11+12=33.
The total is 35, so we have 2 more items to distribute To maintain the p>e>r inequality, the only possible distribution is adding the 2 items to p.
This gives: r=10, e=11, p=14 The sum is $10+11+14=35$ This is the only valid solution.
Q23: If each pen cost ₹20, each ruler cost ₹12 and each eraser cost ₹5, find the minimum amount (in ₹) that Rohan spent for purchasing the items.
Correct Option:
Correct Answer: 455
Explanation: From the given conditions (p > e > r; p, e, r >= 10; p+e+r=35), the only possible combination of items is 14 pens, 11 erasers, and 10 rulers.
Total Cost = (14 pens × ₹20) + (11 erasers × ₹5) + (10 rulers × ₹12) = ₹280 + ₹55 + ₹120 = ₹455.
Q24: What could be the actual number of toys sold?
19
49
91
94
Correct Option: C
Correct Answer: 91
Explanation: Let the actual number of toys be N = 10a+b The mistaken number is N' = 10b+a The stock showed 72 items more, meaning the recorded sale was 72 items less than the actual sale $N - N' = 72$
$(10a+b) - (10b+a) = 72$
$9a - 9b = 72 \implies a - b = 8$.
Since a and b are digits (and a≠0), the only possibilities for (a,b) are (8,0) and (9,1) This means the actual number of toys sold (N) could be 80 or 91 Of the choices given, 91 is a possible value.
Q25: If the faulty calculations show a total sale of ₹1577, what was the actual selling price of each toy?
38
57
75
83
Correct Option: A
Correct Answer: 38
Explanation: From the previous question, the actual number of toys sold (N) was 80 or 91, which means the faulty number sold (N') was 8 or 19 Faulty Sale = Faulty Number × Faulty Price = 1577.
If N' = 8, Price = 1577/8 (not an integer).
If N' = 19, Price = 1577/19 = 83.
So, the faulty price was P' = 83 If P' = 10d+c, then d=8 and c=3.
The actual price is P = 10c+d = 10(3) + 8 = 38.
Q26: A so-called great gambler started playing a card game with a certain amount of money. In the first round he tripled his amount and he gave away p to his friend. In the second round he doubled the amount with him and gave away 3p to his friend. In the third round he quadrupled the amount with him and gave away 2p to his friend and was finally left with no money. If he gave away a total of ₹360 to his friend, then what was the amount of money that he started with (in ₹)?
Correct Option:
Correct Answer: 55
Explanation: 1: Find p Total given away = $p + 3p + 2p = 6p$ $6p = 360 \implies p = 60$. 2: Work backwards.
• Before giving away 2p (₹120) in round 3, he had 0 + 120 = 120.
• This was after quadrupling, so before round 3, he had 120 / 4 = 30.
• Before giving away 3p (₹180) in round 2, he had 30 + 180 = 210.
• This was after doubling, so before round 2, he had 210 / 2 = 105.
• Before giving away p (₹60) in round 1, he had 105 + 60 = 165.
• This was after tripling his initial amount, so he started with 165 / 3 = 55.
Q27: In t minutes, the time would be 8:00 a.m. If 40 minutes ago, the time was 3t minutes past 2:00 a.m., then find the present time.
6:20 a.m.
6:40 a.m.
5:20 a.m.
5:40 a.m.
Correct Option: B
Correct Answer: 6:40 a.m.
Explanation: Let the present time be T.
1 $T + t = 8:00 a.m.
2 $T - 40 \text{ mins} = 2:00 \text{ a.m.} + 3t$
From (1), $T = 8:00 - t$ Substitute this into (2):
$(8:00 - t) - 40 \text{ mins} = 2:00 + 3t$
$7:20 - t = 2:00 + 3t$
$5 \text{ hours } 20 \text{ mins} = 4t$
$320 \text{ mins} = 4t \implies t = 80 minutes.
Present time T = 8:00 a.m - 80 minutes = 6:40 a.m.
Q28: An exam has 120 questions. Each correct answer carries 1 mark. Each wrong answer is penalized by $\frac{1}{3}^{rd}$ of a mark and each unanswered question is penalized by $\frac{1}{6}^{th}$ of a mark. A student who attempted the exam scored 60 marks. The minimum number of answers that the student could have got wrong is
Correct Option:
Correct Answer: 3
Explanation: Let c, w, u be the number of correct, wrong, and unanswered questions.
1 $c+w+u = 120$
2 $c - \frac{w}{3} - \frac{u}{6} = 60 \implies 6c - 2w - u = 360$
Substitute u = 120 - c - w into the second equation:
$6c - 2w - (120 - c - w) = 360$
$7c - w = 480 \implies w = 7c - 480$.
To minimize w, we must minimize c Since w \geq 0, $7c \geq 480 \implies c \geq 68.57$ The minimum integer value for c is 69.
Minimum w = 7(69) - 480 = 483 - 480 = 3.
Q29: Bala had three sons. He had some chocolates which he distributed among them. To his eldest son, he gave 3 more than half the number of chocolates with him. To his second eldest son he gave 4 more than one-third of the remaining chocolates with him. To his youngest son he gave 4 more than one-fourth of the remaining chocolates with him. He was left with 11 chocolates. How many chocolates did he initially have?
180
78
144
120
Correct Option: D
Correct Answer: 78
Explanation: Work backwards from the 11 chocolates left.
• Before Youngest: Let the amount be C3 C3 - (\frac{1}{4}C3 + 4) = 11 \implies \frac{3}{4}C3 = 15 \implies C3 = 20.
• Before Second Eldest: Let the amount be C2 C2 - (\frac{1}{3}C2 + 4) = 20 \implies \frac{2}{3}C2 = 24 \implies C2 = 36.
• Before Eldest (Initial): Let the amount be C1 C1 - (\frac{1}{2}C1 + 3) = 36 \implies \frac{1}{2}C1 = 39 \implies C1 = 78.
Q30: Prakash, Sameer, Ramesh and Tarun have a total of ₹240 with them. Prakash has half the total amount of what the others have. Sameer has one-third of the total amount of what the others have. Ramesh has one-fourth of the total amount of what the others have. Find the amount with Tarun (in ₹)
Correct Option:
Correct Answer: 52
Explanation: Let P, S, R, T be the amounts Total = 240.
• $P = \frac{1}{2}(S+R+T) = \frac{1}{2}(240-P) \implies 3P = 240 \implies P=80$.
• $S = \frac{1}{3}(P+R+T) = \frac{1}{3}(240-S) \implies 4S = 240 \implies S=60$.
• $R = \frac{1}{4}(P+S+T) = \frac{1}{4}(240-R) \implies 5R = 240 \implies R=48$.
• $T = 240 - P - S - R = 240 - 80 - 60 - 48 = 52$.
Q31: There are ten children standing in a line, not all of whom have the same number of chocolates with them. If the first child distributes his chocolates among the remaining nine such that he doubles their respective number of chocolates then he will be left with one chocolate. If the tenth child takes away one chocolate from each of the remaining nine then he will have four chocolates less than the first child initially had. What is the total number of chocolates with the second child to the ninth child?
Correct Option:
Correct Answer: 12
Explanation: Let c1, c2, .. c10 be the chocolates with each child Let S = c2 + .. + c9.
1 First child (c1) gives c2+c3+...+c10 and is left with 1 $c1 = (c2+...+c9) + c10 + 1 \implies c1 = S + c10 + 1$.
2 Tenth child (c10) takes 1 from each of the other 9 New amount is c10+9 $c10 + 9 = c1 - 4 \implies c1 = c10 + 13$.
Equating the two expressions for c1:
$S + c10 + 1 = c10 + 13 \implies S = 12$.
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