Breakup G Strategy: Ratios
The ratio of the number of coins in boxes A and B was 17:7. After 108 coins were shifted from box A to box B, this ratio became 37:20. The number of coins to be further shifted from box A to box B to make this ratio 1:1 is:
[TITA]
Solution & Explanation
Correct Answer: 272
Let the number of coins in Box A be 17x and in Box B be 7x.
After shifting 108 coins:
New number of coins in A = 17x – 108, in B = 7x + 108.
The ratio becomes 37:20, so we set up the equation:
(17x – 108) / (7x + 108) = 37 / 20
Solving this gives x = 76.
Coins in A = 17 * 76 = 1292, and in B = 7 * 76 = 532.
After shifting 108 coins:
A has 1184 coins, and B has 640 coins.
To make the ratio 1:1:
Set up the equation for the 1:1 ratio:
(1184 – y) / (640 + y) = 1
Solving gives y = 272.
Answer 272
Breakup G Strategy | 1.2 / 0.8 Percentage + Profit Loss | Moderate
The monthly sales of a product from January to April were 120, 135, 150 and 165 units respectively. The cost per unit of the product was Rs 240 per unit and a fixed marked price was used for the product in all the four months. However, 20%, 10% and 5% discount on the marked price was offered in January, February, March respectively while no discount were given in April.If the total profit from January to April was Rs 138825, then the marked price (in Rs) is equal to, say:
1.515
2.525
3.520
4.510
Solution & Explanation
Correct Answer: 2
Given: Units sold: January (120), February (135), March (150), April (165)
Cost per unit: Rs 240
Discounts: 20% (January), 10% (February), 5% (March), 0% (April)
Total profit: Rs 138,825
Let M be the marked price.
Revenue for Each Month
January: 120 × 0.80M = 96M
February: 135 × 0.90M = 121.5M
March: 150 × 0.95M = 142.5M
April: 165 × M = 165M
Total Revenue = 96M + 121.5M + 142.5M + 165M = 525M
Total Cost = (120 + 135 + 150 + 165) × 240 = 136800
Profit = Revenue – Cost
138825 = 525M – 136800
525M = 275625
M = 275625 / 525 = 525
The marked price M = Rs 525, so the correct answer is Option 2.
Breakup G Strategy | DPAC Logarithm + Equations | Hard
The sum of all possible real values of x for which
logx-3(x² – 9) = logx-3(x + 1) + 2 is :
1.√33
2.-3
3.(3+√33)/2
4.3
Answer & Explanation
log(x-3) ((x²−9)/(x+1)) = 2
By definition of log :
(x²−9)/(x+1) = (x−3)²
Now equating :
(x²−9)/(x+1) = x²−6x+9
x²−9 = (x²−6x+9)(x+1)
Solve the cubic equation:
x³−6x²+3x+18 = 0
Factor the cubic equation:
(x−3)(x²−3x−6) = 0
Solve x²−3x−6 = 0 using the quadratic formula:
x = (3 ± √33)/2
Sum of solutions:
Sum = (3 + √33)/2
Option 3 is the answer
X Maro G Strategy | Surds and Indices | Moderate
If 12¹²x × 4²⁴x + ¹² × 5²y = 8⁴z × 20¹²x × 2433x-6 , where x, y and z are natural numbers, then x+y+z equals_____
Solution & Explanation
12¹²x × 4²⁴x + ¹² × 5²y = 8⁴z × 20¹²x × 2433x-6
12 = 2² × 3 , 4 = 2² , 8 = 2³ , 20 = 2² × 5 , 243 = 3⁵
Thus expression can be written as :
224x × 312x × 248x+24 × 52y = 212z × 2²⁴x × 512x × 315x – 30
Cancels out the same powers and comparing the powers
12x = 15x – 30 , x = 10
2y = 12x , 2y = 12(10) , y = 60
12 z = 48x + 24 , 12z = 504 , z = 42
x + y + z = 10 + 60 + 42 = 112
Answer 112
Breakup G Strategy | AP , GP + Equation | Moderate
In an arithmetic progression, if the sum of fourth, seventh and tenth term is 99, and the sum of the first fourteen terms is 497, then the sum of first five terms is : [TITA]
Solution & Explanation
Given : Sum of 4th, 7th, and 10th terms = 99
Sum of first 14 terms = 497
Use the formulas for AP
(a + 3d) + (a + 6d) + (a + 9d) = 99
This simplifies to:
a + 6d = 33 (Equation 1)
(14/2) × [2a + 13d] = 497
2a + 13d = 71 (Equation 2)
Solve the system of equations
From Equation 1:
a = 33 – 6d
Substitute this into Equation 2:
2(33 – 6d) + 13d = 71
Simplifying gives:
d = 5, a = 3
The first 5 terms are:
T1 = 3, T2 = 8, T3 = 13, T4 = 18, T5 = 23
Sum of the first 5 terms:
3 + 8 + 13 + 18 + 23 = 65
Answer: 65
Visual lens G Strategy | Number Systems | Easy
The sum of all the digits of the number (10⁵⁰ + 10²⁵ – 123), is
1.212
2.255
3.221
4.324
Solution & Explanation
Correct Answer: 221
The number 10⁵⁰ is a 1 followed by 50 zeros.
The number 10²⁵ is a 1 followed by 25 zeros.
On adding , we have
1000000… upto 24zeros 100000…. upto 25 zeros
When you subtract 123 from this number,
The 25 zeros that follow it are transformed.
The first 25 – 3 = 22 zeros all become 9s.
The last three digits become 1000 – 123 = 877
So number will be
1000 …upto 24 zeros 999……. upto 22 times 877
Now sum of all digits :
1 (the first digit) + 0 (from the 24 zeros)+ 0 (from the 25 zeros )+ (22 × 9) (from the 22 nines)+ (8 + 7 + 7) (from the last three digits) = 1 + 0 + 0 + 198 + 1 = 221
Breakup G Strategy | Average + Percentage + Equations | Moderate
The average salary of 5 managers and 25 engineers in a company is Rs 60000. If each of the manager receive 20% hike in the salary while the salary of the engineer remains unchanged , the average salary of all 30 employees increased by 5%.The average salary, of engineer , in rs , is :
1.45000
2.54000
3.50000
4.40000
Solution & Explanation
Correct Answer: 2 (5400)
Total salary of 30 employees (5 managers and 25 engineers):
30 × 60,000 = 1,800,000
After 20% hike for managers, average salary increases by 5%, so new average salary is:
60,000 × 1.05 = 63,000
Total salary after the hike:
30 × 63,000 = 1,890,000
Before hike: 5M + 25E = 1,800,000
After hike: 6M + 25E = 1,890,000
Solve for M (manager’s salary):
(6M + 25E) – (5M + 25E) = 1,890,000 – 1,800,000
M = 90,000
Solve for E (engineer’s salary):
5 × 90,000 + 25E = 1,800,000
450,000 + 25E = 1,800,000
25E = 1,350,000 Thus E = 54,000
X maro G Strategy | Inequalities | Easy
For real values of x, the range of the function f(x) = (2x – 3)/(2x² + 4x – 6)
1.(- oo , 1/8] U [1 , oo)
2.(-oo , 1/4] U [1/2 , oo)
3.(-oo , 1/8] U [1/2 , oo)
4.(-oo , 1/4] U [1 , oo)
Solution & Explanation
Correct Answer: 3
We can factor the denominator:
2x² + 4x – 6 = 2(x + 3)(x – 1)
So, the function becomes:
f(x) = (2x – 3) / (2(x + 3)(x – 1)) , thus x ≠ -3 , 1
Range of f(x)
f(0) = 1/2
f(1) = not possible
f(2) = 1/10
f(3) = 1/8
Only option 3 includes f(0) and f(3)
Thus option 3 is the answer
Concepts of Algebra | Moderate
f(x) = (x²+ 3x)(x² + 3x + 2), then the sum of all real roots of the equation √f(x) + 1 = 9701, is
1.3
2.6
3.-3
4.-6
Solution & Explanation
Correct Answer: 3
Assume x² + 3x = a , so x² + 3x + 2 = a + 2
Fx = a.(a + 2) = a² + 2a
√fx + 1 = √a² + 2a + 1 = 9701
√(a+1)² = 9701 , a = 9700
Thus x² + 3x = 9700 , x² + 3x – 9700 = 0
Sum of roots = -b/a = -3/1 = -3
Thus option 3 is the answer
Square ±2 G Strategy | Easy
If (x²+ 1/x²) = 25 and x > 0, then the value of (x⁷ + 1/x⁷) is:
1.44856√3
2.44853√3
3.44859√3
4.44850√3
Solution & Explanation
Correct Answer: 44853√3
Given x² + 1/x² = 25 , x > 0
Using identity (a + b)² = a² + b² + 2ab
I. x + 1/x:
(x + 1/x)² = x² + 2 + 1/x² = 25 + 2 = 27
So, x + 1/x = 3√3
II. x³ + 1/x³:
x³ + 1/x³ = (x + 1/x) * (x² + 1/x²) – (x + 1/x)
x³ + 1/x³ = (3√3) * 25 – 3√3 = 72√3
III.x⁴ + 1/x⁴:
x⁴ + 1/x⁴ = (x² + 1/x²)² – 2
x⁴ + 1/x⁴ = 25² – 2 = 625 – 2 = 623
IV.x⁷ + 1/x⁷:
x⁷ + 1/x⁷ = (x³ + 1/x³) * (x⁴ + 1/x⁴) – (x + 1/x)
x⁷ + 1/x⁷ = (72√3) * 623 – 3√3 = 44853√3
Visual lens G Strategy | Set theory + Maximum/Minimum | Moderate
In a class of 150 students, 75 students chose physics, 111 students chose mathematics and 40 students chose chemistry. All students chose at least one of the three subjects and at least one student chose all three subjects. The number of students who chose both physics and chemistry is equal to the number of students who chose both chemistry and mathematics, and this is equal to half the number of students who chose both mathematics and physics. What is the maximum possible number of students who chose physics but not mathematics?
1.40
2.35
3.55
4.30
Solution & Explanation
Given Information: Physics students = 75, Mathematics students = 111 , Chemistry students = 40
Assume: x – number of students who chose both Physics and Mathematics , y – students who chose both Mathematics and Chemistry , z – student who chose both Physics and Chemistry , t – number of students who chose all three subjects
And z = y , y = x/2 (Given)
Using Set formula (the total number of students who chose at least one subject)
|P ∪ M ∪ C| = |P| + |M| + |C| − |P ∩ M| − |M ∩ C| − |P ∩ C| + |P ∩ M ∩ C|
Substitute the values:
150 = 75 + 111 + 40 – x – y – z + t
Now, substitute z = y and y = x/2:
150 = 75 + 111 + 40 – x – (x/2) – (x/2) + t
150 = 226 – x – x + t
150 = 226 – 2x + t
2x – 76 = t (Equation 1)
We want to maximize 75 – x , (students who chose only Physics but not mathematics). Using options :
1.40 , 75 – x = 40 , x = 35 , t can not be negative
2.35 , 75 – x = 35 , x = 40 , t = 4
3.55 , 75 – x = 55 , x= 20 , t can not be negative
4.30 , 75 – x = 30 , x = 45 , t = 14
Since 75 – x is maximum at option 2 thus option 2 is the answer
Visual lens G Strategy (Concept of similarities) | Easy
In a triangle ABC, AB=AC=12 cm and D is a point on side BC such that AD= 8 cm. If AD is extended to point E such that angle ACB = angle AEB , then length in cm of AE is :
1.16
2.18
3.14
4.20
Answer & Explanation
Based on the information in the question triangle ABD ≈ triangle AEB , Thus :
AB/AE = AD/AB ,
12/AE = 8/12 = 2/3
Only option 2 satisfies the condition. Thus option 2 is the answer
Visual G Strategy | Triplets + Area of Triangles | Moderate
A triangle ABC is formed with AB = AC = 50 cm and BC = 80 cm. Then, the length, in cm , of all three altitudes of the triangle ABC is….. …
Solution & Explanation
Give AB = AC = 50 cm , BC = 80 cm
The altitude from A to BC
Half of BC: 80 / 2 = 40 cm
Use Pythagoras’ theorem to find altitude h:
h² + 40² = 50²
h² + 1600 = 2500
h² = 900 , h = 30 cm
Altitude from A to BC = 30 cm
The altitudes from B and C
Area of the triangle using the altitude from A:
Area = (1/2) × 80 × 30 = 1200 cm²
Using area to find the altitude from B:
1200 = (1/2) × 50 × h
h = 1200 / 25 = 48 cm
Altitude from B to AC = 48 cm
Altitude from C to AB = 48 cm
Sum = 30 + 48 + 48 = 126
Answer 126
Visual G Strategy | Cyclic quadrilateral+ Area of trapezium | Moderate
ABCD is a trapezium in which AB is parallel to DC and AD is perpendicular to AB and AB = 3DC. If a circle inscribed in a trapezium touching all the sides has a radius 3cm , then the area in sq cm , of the trapezium is :
1.48
2.36√2
3.54
4.30√3
Solution & Explanation
Given:
Let the trapezium be ABCD.
AB || DC (The parallel sides are the bases).
AD perpendicular to AB (The trapezium is a right-angle trapezium). AB = 3DC. If AB = x , then DC = 3x
The circle inscribed in the trapezium has a radius r = 3cm
Height = 2r = 2 × 3 = 6cm
For cyclic quadrilateral Sum of parallel sides= Sum of non-parallel sides
AB + DC = AD + BC
3x + x = 6 + BC
BC = 4x – 6 —– I
In Triangle AEB with base BE , height CE and hypotenuse BC
BE = DC – AB = 3x – x = 2x , CE = 6cm , BC = 4x – 6
Using Pythagoras:
(2x)² + (6)² = (4x – 6)²
On solving we have x = 4cm
Area of trapezium = ½ * (AB + DC) × AD
= ½ .(4x) .6 = 48cm² , This option 1 is the answer
Breakup G Strategy: Numbers + Maximum/Minimum | Hard
Let p, q and r be three natural numbers such that their sum is 900, and r is a perfect square whose values lies between 150 and 500.If p is less than 0.3q and not more than 0.7q, then the sum of the maximum and minimum possible values of p is: _____
Solution & Explanation
Correct Answer: 397
Give p + q + r = 900 , 0.3q ≤ p ≤ 0.7q
r lies between 150 to 500 and is a perfect square
So r = 169 , 196 , 225 , 256 , 289 , 324 , 361 , 400 , 441 , 484
For maximum p (0.7q) , r should be minimum, r = 169
0.7q + 1q + 169 = 900 , q = 430 so p = 0.7 × 430 = 301
For minimum p (0.3q) , r should be maximum r = 484
0.3q + 1q + 484 = 900 , q = 320 , so p = 0.3 × 320 = 96
Pmax + P min = 301 + 96 = 397
Answer 397
Concepts of Number system + Maximum/Minimum | Hard
For a 4 digit number greater than 1000 , sum of the digits in thousands , hundreds and tens places is 15.And the sum of digits in hundreds , tens and units place is 16.Also the digits in tens place is 6 more than the digit in the unit place. The difference between the largest possible and smallest possible value of the number is :
1.735
2.3289
3.4078
4.811
Solution & Explanation
Given conditions for a 4-digit number:
a + b + c = 15 —– I
b + c + d = 16 —– II
c = d + 6 ——- III
Solving the equations:
b = 10 – 2d
a = d – 1
Now using values for d:
If d = 3:
a = 2, b = 4, c = 9, number = 2493
If d = 2:
a = 1, b = 6, c = 8, number = 1682
Largest number = 2493, Smallest number = 1682
Difference = 2493 – 1682 = 811
Answer 811
Breakup G Strategy | Linear Equation + Maximum/Minimum | Moderate
In a school with 1500 students , each students chooses one of the streams out of science , arts and commerce by paying a fee of rs 1100 , 1000 and 800 respectively. The total fees paid by all the students is RS. 15,50000.If the number of science students is not more than the number of arts students , then the maximum possible number of science students in school is : [TITA] _______
Solution & Explanation
Correct Answer: 700
Given S – science students , A – arts students , C – commerce students
S + A + C = 1500 —- I
1100S + 1000A + 800C = 1550000 becomes
11S + 10A + 8C = 15500 ——- II
Where S ≤ A , since we have to maximise S this S = A
Multiplying I by 8 :
8S + 8A + 8C = 12000 —— III
Subtracting III from I
3S + 2A = 3500 , Since S = A
5S = 3500 , S = 700
Answer : 700
Breakup G Strategy | Average + Percentage + Equations | Moderate
The average salary of 5 managers and 25 engineers in a company is Rs 60000. If each of the manager receive 20% hike in the salary while the salary of the engineer remains unchanged , the average salary of all 30 employees increased by 5%.The average salary, of engineer , in rs , is :
1.45000
2.54000
3.50000
4.40000
Solution & Explanation
Correct Answer: 2 (5400)
Total salary of 30 employees (5 managers and 25 engineers):
30 × 60,000 = 1,800,000
After 20% hike for managers, average salary increases by 5%, so new average salary is:
60,000 × 1.05 = 63,000
Total salary after the hike:
30 × 63,000 = 1,890,000
Before hike: 5M + 25E = 1,800,000
After hike: 6M + 25E = 1,890,000
Solve for M (manager’s salary):
(6M + 25E) – (5M + 25E) = 1,890,000 – 1,800,000
M = 90,000
Solve for E (engineer’s salary):
5 × 90,000 + 25E = 1,800,000
450,000 + 25E = 1,800,000
25E = 1,350,000 Thus E = 54,000
Breakup G Strategy: Numbers + Maximum/Minimum | Hard
Let p, q and r be three natural numbers such that their sum is 900, and r is a perfect square whose values lies between 150 and 500.If p is less than 0.3q and not more than 0.7q, then the sum of the maximum and minimum possible values of p is: _____
Solution & Explanation
Correct Answer: 397
Give p + q + r = 900 , 0.3q ≤ p ≤ 0.7q
r lies between 150 to 500 and is a perfect square
So r = 169 , 196 , 225 , 256 , 289 , 324 , 361 , 400 , 441 , 484
For maximum p (0.7q) , r should be minimum, r = 169
0.7q + 1q + 169 = 900 , q = 430 so p = 0.7 × 430 = 301
For minimum p (0.3q) , r should be maximum r = 484
0.3q + 1q + 484 = 900 , q = 320 , so p = 0.3 × 320 = 96
Pmax + P min = 301 + 96 = 397
Answer 397
LCM G Strategy Concept of Time and Work | Hard
Teams A, B, and C consist of five, eight, and ten members, respectively such that the members within each team is equally productive. Working separately teams A , B and C can complete certain job in 40 hours , 50 hours and 4 hours respectively Two members from team A, three members from team B, and one member from team C together starts the work and the member from team C left the work after 23 hours. The number of additional member(s) required from team B , that would be required to replace the member from team C , to finish the job in next one hour is :
1.4
2.2
3.3
4.1
Solution & Explanation
Correct Answer: 2
Total work that has to be completed (LCM of 40,50,4) = 200 units
Efficiencies
Member of team A = 200/40 = 5units per hour , so 1 member (5/5 = 1 unit work per hour)
Member of team B = 200/50 = 4 units per hour , so 1 member (4/8 = 0.5 unit work per hour)
Member of team C = 200/4 = 50 units per hour , so 1 member (50/10 = 5 unit work per hour)
Work Done in 23 hours:
2 + 1.5 + 5 = 8.5 units/hour
Work Done in 23 hours: 8.5 × 23 = 195.5 units
Remaining work = 200 – 195.5 = 4.5 units
Now C left the work , A and B has to complete the remaining work in 1 hour :
2 (1) + (3 + x) × 0.5 = 4.5
On solving x = 2
Thus option 2 is the answer
Breakup G Strategy: TSD + Equation | Moderate
Rahul starts on his journey at 5pm at a constant speed so that he reaches his destination at 11pm same day . However , on his way , he stops for 20 minutes and after that, increases his speed by 3 km per hour to reach on time. If he had to stop for 10 minutes more , he would have had to increase his speed by 5km per hour to reach on time.His initial speed in km per hour was :
1.18
2.15
3.12
4.20
Solution & Explanation
Correct Answer: 2
Let Rahul’s initial speed be V km/h.
The total distance D is D = V × 6.
When he stops for 20mins (6-⅓ = 17/3) , t – l time
6V = (v × t) + (v + 3) × (17/3 – t)
On solving we have v/3 = 17 – 3t , v = 51 – 9t —- I
When he stops for 30 mins (6-½) = 11/2
6v = (v × t) + (v + 5) (11/2 – t)
On solving V/2 = 55/2 – 5t , v = 55 – 10t —— II
On solving I and II , we have t = 4 , then v = 15km/hr
Thus option 2 is the answer
Breakup G Strategy: Time Speed + Percentage | Moderate
Ankita walks from A to C through B, and runs back through the same route at a speed that is 40% more than her walking speed. The total time taken is 3 hours 30 minutes to walk from B to C as well as to run from B to A. The total time, in minutes, required to walk from A to B and run from B to C is: [TITA]
Solution & Explanation
Correct Answer: 444
Assume walking speed = s , Running speed = 1.4s
Distance from A to B = d1 , Distance from B to C = d2
Now based on the information given in the question:
d2/v = 3.5 hrs , d1/1.4v = 3.5 hrs ,
d1/v = 3.5 × 1.4
Now total time required to walk from A to B and run from B to C.
Total time = d1/v + d2/1.4v = (3.5 × 1.4) + (3.5/1.4) =
4.9 hrs + 2.5 hrs = 7.4 hrs = 7hrs 24 mins
7hrs 24 mins = (7 × 60) + 24 = 444 mins
Answer 444
Breakup G Strategy : Mixture + Equation | Moderate
Vessels A and B contain 60 litres of alcohol and 60 litres of water, respectively. A certain volume is taken out from A and poured into B. After mixing well, an equal volume is taken out from B and poured into A. If the ratio of alcohol and water in A is 15:4, then the volume, in litres, initially taken out from A, is [TITA]
Solution & Explanation
Correct Answer: 16
Given vessels A and B contain 60 litres of alcohol and 60 litres of water, respectively.
Assum x litres of alcohol from A is transferred to B
After the first transfer (from A to B):
Vessel A: 60 – x liters of alcohol, 60 liters of water.
Vessel B: x liters of alcohol, 60 liters of water.
After the second transfer (from B to A):
Alcohol transferred back to A: x * (x / (60 + x)).
Water transferred to A: x * (60 / (60 + x)).
Alcohol and water in A after the second transfer:
Alcohol in A: 60 – x + (x² / (60 + x)).
Water in A: 60 + (60x / (60 + x)).
Set up ratio equation (15:4):
(60 – x + (x² / (60 + x))) / (60 + (60x / (60 + x))) = 15 / 4
Solve for x: x = 16 liters
LCM G Strategy | Time and Work | Easy
The rate of water flow through three pipes A, B and C are in the ratio 4 : 9 : 36. An empty tank can be filled by pipe A in 15 hours. If all pipes are used simultaneously to fill up this empty tank, the time, in minutes, taken to fill up the tank entirely tank completely is nearest to :
1.71
2.78
3.76
4.73
Solution & Explanation
Pipe A fills the tank in 15 hours.
Pipe A’s rate = 1/15 of the tank per hour.
Let the flow rates be:
Pipe A’s rate = 4x , Pipe B’s rate = 9x ,Pipe C’s rate = 36x
From 4x = 1/15, we find x = 1/60. Now putting x = 1/60 , we have pipe A , B , C rates as 4/60 , 9/60 , 36/60 respectively
Now, add the rates:
Total rate = 4/60 + 9/60 + 36/60 = 49/60
The time to fill the tank is the reciprocal of the total rate:
Time = 1 / (49/60) = 60 / 49 hours.
Convert this to minutes:
Time in minutes = (60 / 49) × 60 ≈ 73.47 minutes ≈ 73
