# Mixed Quantitative Aptitude Questions Set 12

1. A sum of money doubles itself at Simple Interest in 10 years at some rate of interest. What interest will a sum of Rs 10,000 pay after 1 year at the same rate of interest?
A) Rs 1,000
B) Rs 1,200
C) Rs 800
D) Rs 1,500
E) None of these
Option A
Explanation
:
Here A = 2P => P = SI
P = (P*r*10)/100 =>
r = 10%
=> Interest  Rs 1000
2. Mr. X buys a mobile at Rs. 10,000 whose value depreciates at the rate 5% p.a. After 1 year of buying the phone he had to spend Rs 2,000 on its repair. After further 1 year he sold the mobile for Rs 8,000. Find the profit/loss % that he incurred in this transaction.
A) 33 (1/3) % Loss
B) 33 (1/3) % Profit
C) 50% Loss
D) 20% Loss
E) None of these
Option A
Explanation
:
CP = 10000 + 2000 = 12000 ; SP = Rs 8000 ; find loss %
3. Mr. X invested in a saving scheme which offers an interest rate of 8% compounded annually. The maturity period of the scheme is 3 years. But 1 year before the maturity period Mr. X withdrew his amount. For this premature withdrawal, the issuer bank deducted 1% of the total amount payable to Mr. X for that period. If after deduction Mr. X got Rs 11,54,736 then find his total investment.
A) Rs9,50,000
B) Rs 9,00,000
C) Rs 10,00,000
D) Rs 10,50,000
E) None of these
Option C
Explanation
:
Rs 11,54,736 is 99% of the total amount without deduction.(If 1 % was not deducted)
100% = 1154736*100/99 = Rs. 11,66,400
Using this as amount find the principle for 2 years and rate of 8%
P = 1166400/1.1664 = Rs 10,00,000
4. A boat travels a distance downwards from Point A to Point B and then moves upwards from point B back to point A. The speed of boat in still water is 5 kmph while the normal speed of stream is 1 kmph. Find the time taken by the boat to travel the whole distance to and fro if the speed of the stream doubles in the return journey (From B to A). The distance between A to B is 10 km
A) 8 hours
B) 9 hour 30 minutes
C) 5 hours
D) 7 hours
E) 8 hour 40 minutes
Option C
Explanation
:
T = 10/[(5+1)] + 10/[(5-2)]
T = 5 hours
5. Train X travelling at a speed of 72 kmph crosses a platform of equal length in 15 seconds more than it takes to cross a signal post. In how much time will another train Y of length 450 meters travelling at a speed of 36 kmph and Train X cross each other while moving towards each other?
A) 30 second
B) 25 second
C) 20 second
D) 15 second
E) None of these
Option B
Explanation
:
Let  length of train X = Length of platform = X kmph.
2X/20 – X/20 = 15 [72 kmph = 20 m/s]
=> X = 300 m
T=(300+450)/(20+10) [m/s equivalent of kmph]
6. A dishonest shopkeeper Mr.X marks his good 10% above the Cost Price and also uses a false weight of 900 grams instead of 1 kg. One day his rival shopkeeper replaced this false weight of 900 grams with a weight of 1100 grams, so that Mr. X incurs loss. Find the %  Gain/Loss that Mr. X will incur after the weight is replaced.
A) 5 % Profit
B) 5 % Loss
C) 10% Profit
D) 10 % Loss
E) No Profit No Loss
Option E
Explanation
:
1000/1100 = (100+G)/(100+10) => G = 0%
7. Mr. X can do a work in few days. Mr. Y who is twice as efficient as Mr. X can do the same work in 10 days. Find the number of days in which Mr. X, Mr. Y and Mr. Z will complete the work together if Mr. Z is half as efficient as Mr. X
A) 15 days
B) 40/7 Days
C) 36/7 Days
D) 8 days
E) None of these
Option B
Explanation
:
Y-> 10 days ;
X -> 20 days (Y is twice efficient as X)
Z =  40 days
1/days = 1/10 + 1/20  + 1/40
=>Days = 40/7
8. A man and his wife appear in an examination for two vacancies for the same post. The probability of husband′s selection is (1/6) and the probability of wife’s selection is (1/4). What is the probability that only one of them is selected?
A) 1/3
B) 1/4
C) 1/5
D) 1/6
E) None of these
Option A
Explanation
:
P(Man selected) = 1/6 ; P(Man not selected) = 1 – 1/6  = 5/6
P(Wife selected)= 1/4 ; P(Wife not selected) = 3/4
Required Probability = (Man selected & Wife not selected) OR (Wife selected & Man not selected)
= (1/6 *3/4 ) + ( 1/4 *5/6 )
= 1/3
9. A square field which was newly built incurred a total cost of Rs 16,000 in fencing at a rate of Rs 50 per meter. But later it was realized that the field has been wrongly constructed, and the new dimension of field should be 100 meter * 80 meter. What is the minimum possible cost that will be incurred for fencing the extra part of field at the same rate.
A) Rs 8,000
B) Rs 5,000
C) Rs 6,000
D) Rs 10,000
E) Rs 18,000
Option C
Explanation
:
Perimeter of field  = 16000/50 = 320
=>side of square = 80 meter
Now the total cost will be minimum, when one of the side of  the field is demolished and exteneded by 20 m on its length to get the total length of 100 m, and then the 80 m demolished boundary is constructed.
So the total length now to be fenced = 20+80+20 meter = 120 meter
Total cost =120*50 = Rs 6,000
10. The age of Mr X will double in 30 years. B is the sister of Mr. X and is 5 year younger to him. B is married to Mr. Z who is 5 year elder to Mr. X What will be the average Age of B and Mr. Z after 10 years?
A) 30
B) 35
C) 40
D) 45
E) None of these
Option C
Explanation
:
Age of X = x
x+30 = 2x
=>x = 30
B = 25 ; Z = 35
After 10 years total age = 25+10 +35+10 = 80
Average = 40 years

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