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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections (15): Read the following information carefully and answer the given questions.
Income of Sarika and Raju are in the ratio 27 : 22 and ratio of their expenditure is 127:102 respectively. Sarika saves Rs._____ and Raju save Rs._____. Manoj spends 50% of his income for rent, food and medicine. Amount he spent for medicine is Rs.5000 which is twoninth of the remaining income. Income of Sarika is 75% of Manoj’s income.
Which of the following satisfy the two blanks given in the questions?
I)Rs.2000, Rs.2000
II)Rs.5000, Rs.3000
III)Rs.7000, Rs.6000Correct
Answer 1) Only I
Explanation:
Let us take income of Manoj be x
Amount spent for medicine = 2/9 * (x50/100*x) = 5000
= > 2/9 * (50x/100) = 5000
= > 2/9 * x/2 = 5000
= > x/9 = 5000
= > x = 45000
Income of Sarika = 45000 * 75/100
= 33750
Income of Raju = 33750/27 * 22
= 27500
From I:
According to the question,
= > (33750 – 2000)/(275002000)
= > 31750/25500
= > 127:102
This satisfies the given condition.
From II:
According to the question,
= > (31750 – 5000)/(275003000)
= > 26750/24500
= > 107:98
This does not satisfy the given condition.
From III:
According to the question,
= > (31750 – 7000)/(275006000)
= > 24750/21500
= > 99:86
This does not satisfy the given condition.Incorrect
Answer 1) Only I
Explanation:
Let us take income of Manoj be x
Amount spent for medicine = 2/9 * (x50/100*x) = 5000
= > 2/9 * (50x/100) = 5000
= > 2/9 * x/2 = 5000
= > x/9 = 5000
= > x = 45000
Income of Sarika = 45000 * 75/100
= 33750
Income of Raju = 33750/27 * 22
= 27500
From I:
According to the question,
= > (33750 – 2000)/(275002000)
= > 31750/25500
= > 127:102
This satisfies the given condition.
From II:
According to the question,
= > (31750 – 5000)/(275003000)
= > 26750/24500
= > 107:98
This does not satisfy the given condition.
From III:
According to the question,
= > (31750 – 7000)/(275006000)
= > 24750/21500
= > 99:86
This does not satisfy the given condition. 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirections (15): Read the following information carefully and answer the given questions.
A Jar contains a mixture of Milk and Water 18 and 12 Liters respectively. When _____ liter of the mixture is taken out and replaced with the same quantity of Water, then the ratio of Milk and Water becomes 2:3. The quantity of Water in final Mixture _____ is?
Which of the following satisfy the two blanks given in the questions?
I)10 litres, 18 litres
II)10 litres, 28 litres
III)20 litres, 36 litresCorrect
Answer a) Only (I)
Explanation:
From I:
Let x litre mixture is taken out,
= >(18x*18/30)/(12 x*12/30+x)
Now putting, x = 10
= >(1810*18/30)/(1210*12/30+10)
= >12/18
= >2/3
Now,
Water in final mixture = 12+ W/ M*10
18=12+ W/M *10
6/10= W/M
W/M= 3/5
It satisfies the condition (I).
From II:
Let x litre mixture is taken out,
= >(18x*18/30)/(12 x*12/30+x)
Now putting, x = 10
= >(1810*18/30)/(1210*12/30+10)
= >12/18
= >2/3
Now,
Water in final mixture = 12+ W/ M*10
28=12+ W/M *10
16/10= W/M
W/M= 8/5
It does not satisfy condition (II)
From III:
Let x litre mixture is taken out,
= >(18x*18/30)/(12 x*12/30+x)
Now putting, x = 20
= >(1820*18/30)/(1220*12/30+10)
= >6/14
= >3/7
Now,
Water in final mixture = 12+ W/ M*10
36=12+ W/M *10
24/10= W/M
W/M= 12/5
It does not satisfy the condition (III)Incorrect
Answer a) Only (I)
Explanation:
From I:
Let x litre mixture is taken out,
= >(18x*18/30)/(12 x*12/30+x)
Now putting, x = 10
= >(1810*18/30)/(1210*12/30+10)
= >12/18
= >2/3
Now,
Water in final mixture = 12+ W/ M*10
18=12+ W/M *10
6/10= W/M
W/M= 3/5
It satisfies the condition (I).
From II:
Let x litre mixture is taken out,
= >(18x*18/30)/(12 x*12/30+x)
Now putting, x = 10
= >(1810*18/30)/(1210*12/30+10)
= >12/18
= >2/3
Now,
Water in final mixture = 12+ W/ M*10
28=12+ W/M *10
16/10= W/M
W/M= 8/5
It does not satisfy condition (II)
From III:
Let x litre mixture is taken out,
= >(18x*18/30)/(12 x*12/30+x)
Now putting, x = 20
= >(1820*18/30)/(1220*12/30+10)
= >6/14
= >3/7
Now,
Water in final mixture = 12+ W/ M*10
36=12+ W/M *10
24/10= W/M
W/M= 12/5
It does not satisfy the condition (III) 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections (15): Read the following information carefully and answer the given questions.
Train M started from station P towards station Q. At the same time, train N started from station R towards station Q. All the three stations are in a straight line such that station Q is between station P and station R. Station Q is equidistant from station P and station R. Distance between station P and station Q is 440 Km. If the speed of train M and train N is ______ Km/hr and ______ Km/hr respectively, the distance between both the trains after five hours is 580 km.
Which of the following satisfy the two blanks given in the questions?
I)32 km/hr, 30 km/hr
II)40 km/hr, 20 km/hr
III)25 km/hr, 35 km/hrCorrect
Answer d) Only II and III
Explanation:
From I:
Distance between station P and station R = 440 x 2 = 880 Km
Distance travelled by train M in 5 hours = 32 x 5 = 160 Km
Distance travelled by train N in 5 hours = 30 x 5 = 150 Km
Required distance = 880 – 160 – 150
= 880 – 310
= 570 Km
This does not satisfy the given condition.
From II:
Distance between station P and station R = 440 x 2 = 880 Km
Distance travelled by train M in 5 hours = 40 x 5 = 200 Km
Distance travelled by train N in 5 hours = 20 x 5 = 100 Km
Required distance = 880 – 200 – 100
= 880 – 300
= 580 Km
This satisfies the given condition.
From III:
Distance between station P and station R = 440 x 2 = 880 Km
Distance travelled by train M in 5 hours = 25 x 5 = 125 Km
Distance travelled by train N in 5 hours = 35 x 5 = 175 Km
Required distance = 880 – 125 – 175
= 880 – 300
= 580 Km
This satisfies the given condition.Incorrect
Answer d) Only II and III
Explanation:
From I:
Distance between station P and station R = 440 x 2 = 880 Km
Distance travelled by train M in 5 hours = 32 x 5 = 160 Km
Distance travelled by train N in 5 hours = 30 x 5 = 150 Km
Required distance = 880 – 160 – 150
= 880 – 310
= 570 Km
This does not satisfy the given condition.
From II:
Distance between station P and station R = 440 x 2 = 880 Km
Distance travelled by train M in 5 hours = 40 x 5 = 200 Km
Distance travelled by train N in 5 hours = 20 x 5 = 100 Km
Required distance = 880 – 200 – 100
= 880 – 300
= 580 Km
This satisfies the given condition.
From III:
Distance between station P and station R = 440 x 2 = 880 Km
Distance travelled by train M in 5 hours = 25 x 5 = 125 Km
Distance travelled by train N in 5 hours = 35 x 5 = 175 Km
Required distance = 880 – 125 – 175
= 880 – 300
= 580 Km
This satisfies the given condition. 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirections (15): Read the following information carefully and answer the given questions.
Tara bought two articles at same price. She sold one at a profit of 20% and other at a loss of 10%. From that amount she bought one laptop and marked its price _____% more than cost price and allowed a discount of 10% on the Marked price. If the cost price of each article was Rs.____, and the overall percent profit earned by Tara is 18.125%.
Which of the following satisfy the two blanks given in the questions?
I)25%, Rs.4800
II)20%, Rs.5000
III)25%, Rs.6000Correct
Answer d) Only I and III
Explanation:
From I:
Cost price of the two articles for Tara = 4800 x 2 = Rs.9600
Selling price of two articles = 4800 x 120/100 + 4800 x 90/100
= 5760 + 4320
= Rs.10080 = cost price of the laptop
Marked price of the laptop = 10080 x 125/100 = Rs.12600
Selling price of the laptop = 12600 x 90/100 = Rs.11340
% profit = [(11340 – 9600)/9600] x 100
= (1740/9600) x 100
= 18.125%
This satisfies the given condition.
From II:
Cost price of the two articles for Tara = 5000*2 = Rs. 10000
Selling price of two articles = 5000 x 120/100 + 5000 x 90/100
= 6000 + 4500
= Rs.10500 = cost price of the laptop
Marked price of the laptop = 10500 x 120/100 = Rs.12600
Selling price of the laptop = 12600 x 90/100 = Rs.11340
% profit = [(11340 – 10500)/10500] x 100
= 840/10500 x 100
= 8%
This does not satisfy the given condition.
From III:
Cost price of the two articles for Tara = 6000 x 2 = Rs.12000
Selling price of two articles = 6000 x 120/100 + 6000 x 90/100
= 7200 + 5400
= Rs.12600 = cost price of the laptop
Marked price of the laptop = 12600 x 125/100 = Rs.15750
Selling price of the laptop = 15750 x 90/100 = Rs.14175
% profit = (14175 – 12000)/12000 x 100
= 2175/12000 x 100
= 18.125%
This satisfies the given condition.Incorrect
Answer d) Only I and III
Explanation:
From I:
Cost price of the two articles for Tara = 4800 x 2 = Rs.9600
Selling price of two articles = 4800 x 120/100 + 4800 x 90/100
= 5760 + 4320
= Rs.10080 = cost price of the laptop
Marked price of the laptop = 10080 x 125/100 = Rs.12600
Selling price of the laptop = 12600 x 90/100 = Rs.11340
% profit = [(11340 – 9600)/9600] x 100
= (1740/9600) x 100
= 18.125%
This satisfies the given condition.
From II:
Cost price of the two articles for Tara = 5000*2 = Rs. 10000
Selling price of two articles = 5000 x 120/100 + 5000 x 90/100
= 6000 + 4500
= Rs.10500 = cost price of the laptop
Marked price of the laptop = 10500 x 120/100 = Rs.12600
Selling price of the laptop = 12600 x 90/100 = Rs.11340
% profit = [(11340 – 10500)/10500] x 100
= 840/10500 x 100
= 8%
This does not satisfy the given condition.
From III:
Cost price of the two articles for Tara = 6000 x 2 = Rs.12000
Selling price of two articles = 6000 x 120/100 + 6000 x 90/100
= 7200 + 5400
= Rs.12600 = cost price of the laptop
Marked price of the laptop = 12600 x 125/100 = Rs.15750
Selling price of the laptop = 15750 x 90/100 = Rs.14175
% profit = (14175 – 12000)/12000 x 100
= 2175/12000 x 100
= 18.125%
This satisfies the given condition. 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections (15): Read the following information carefully and answer the given questions.
P starts a company and after 3 months Q also joins the company. The initial investment of P and Q is in the ratio of 3:2, respectively. P and Q receives Rs. _____ and Rs. _____ respectively as profit after completion of one year of the company.
Which of the following satisfy the two blanks given in the questions?
I)Rs.3000, Rs.1000
II)Rs.1920, Rs.960
III)Rs.2400, Rs.1200Correct
Answer e) Only II and III
Explanation
The ratio of profit share of P and Q is
P: Q = 3 × 12: 2 × (12 – 3) = 2: 1
From I:
The profit ratio of P and Q = 3000: 1000
= 3: 1
This doesn’t satisfy the given condition.
From II:
The profit ratio of P and = 1920: 960
= 2: 1
This satisfies the given condition.
From III:
The profit ratio of P and Q = 2400: 1200
=2:1
This satisfies the given condition.Incorrect
Answer e) Only II and III
Explanation
The ratio of profit share of P and Q is
P: Q = 3 × 12: 2 × (12 – 3) = 2: 1
From I:
The profit ratio of P and Q = 3000: 1000
= 3: 1
This doesn’t satisfy the given condition.
From II:
The profit ratio of P and = 1920: 960
= 2: 1
This satisfies the given condition.
From III:
The profit ratio of P and Q = 2400: 1200
=2:1
This satisfies the given condition. 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeQ(610): Study the data given below and answer the following questions. Number of cars sold by 6 companies during 2 years :
The number of cars sold by F increased by what percent from 2017 to 2018 ?Correct
Answer – 1) 21.5%
Explanation –
(460380)/380 ×100
21.5%Incorrect
Answer – 1) 21.5%
Explanation –
(460380)/380 ×100
21.5% 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeQ(610): Study the data given below and answer the following questions. Number of cars sold by 6 companies during 2 years :
The number of cars sold by company A and B increased by 10% and 15% respectively from 2016 to 2017. What is the total number of cars sold by company A and B together in years 2016, 2017 and 2018 ?Correct
Answer – 2) 2300
Explanation –
Cars sold by A in 2016 = 440 ×100/110 = 400
Cars sold by B in 2016 = 460 ×100/115 =400
Cars sold by A in 2016,2017 and 2018 = 400+ 440 + 320 = 1160
Cars sold by B in 2016, 2017 and 2018 = 400 + 460 + 280 1140
Total cars sold by A and B during 3 years = 1160 + 1140 = 2300Incorrect
Answer – 2) 2300
Explanation –
Cars sold by A in 2016 = 440 ×100/110 = 400
Cars sold by B in 2016 = 460 ×100/115 =400
Cars sold by A in 2016,2017 and 2018 = 400+ 440 + 320 = 1160
Cars sold by B in 2016, 2017 and 2018 = 400 + 460 + 280 1140
Total cars sold by A and B during 3 years = 1160 + 1140 = 2300 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeQ(610): Study the data given below and answer the following questions. Number of cars sold by 6 companies during 2 years :
What is the average number of cars sold per year ?Correct
Answer – 5) 2370
Explanation –
Cars sold in year 2017 = 440 + 460 +340 + 310+290+380 = 2220
Cars sold in year 2018 = 320 + 280+ 450 + 530 + 480+ 460= 2520
Average = (2220+ 2520) /2 = 2370Incorrect
Answer – 5) 2370
Explanation –
Cars sold in year 2017 = 440 + 460 +340 + 310+290+380 = 2220
Cars sold in year 2018 = 320 + 280+ 450 + 530 + 480+ 460= 2520
Average = (2220+ 2520) /2 = 2370 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeQ(610): Study the data given below and answer the following questions. Number of cars sold by 6 companies during 2 years :
The percentage sale increase of which car is maximum in year 2018 with respect to 2017 ?
Correct
Answer – 2) D
Explanation –
C – 110/340
D – 220/310
E – 190/290
Clearly 220/310 is the maximum value.Incorrect
Answer – 2) D
Explanation –
C – 110/340
D – 220/310
E – 190/290
Clearly 220/310 is the maximum value. 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeQ(610): Study the data given below and answer the following questions. Number of cars sold by 6 companies during 2 years :
The year and the name of the comany which sold minimum cars –Correct
Answer – 3) 2018 , Company B
Explanation –
280 cars – the minimum cars sold by company B in 2018.Incorrect
Answer – 3) 2018 , Company B
Explanation –
280 cars – the minimum cars sold by company B in 2018.
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