Quantitative Aptitude - Arithmetic Ability Questions

Q:

Along a yard 225 metres long, 26 trees are palnted at equal distances, one tree being at each end of the yard. What is the distance between two consecutive trees

A) 8 B) 9
C) 10 D) 11
 
Answer & Explanation Answer: B) 9

Explanation:

26 trees have 25 gaps between them,
Required distance (225/25) = 9

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Filed Under: Simplification

6 8312
Q:

Find the C.I. on Rs. 15,225 for 9 months at 16% per annum compounded quarterly ?

A) Rs. 1911 B) Rs. 1909
C) Rs. 1901 D) Rs. 1907
 
Answer & Explanation Answer: C) Rs. 1901

Explanation:

P = Rs. 15225, n = 9 months = 3 quarters, R = 16% p.a. per quarter.

 

Amount = 15225x1+41003

 

= (15225 x 26/25 x 26/25 x 26/25) = Rs. 17126.05

 

=> C.I. = 17126 - 15625 = Rs. 1901.05.

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Filed Under: Compound Interest
Exam Prep: GATE , CAT , Bank Exams , AIEEE
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10 8311
Q:

On a ruler's tombstone, it is said that one sixth of his life was spent in childhood and one twelfth as a teenager. One seventh of his life passed between the time he became an adult and the time he married; five years later, his son was born. Alas, the son died four years before he did. He lived to be twice as old as his son did. How old did the ruler live to be ?

A) 84 years B) 72 years
C) 82 years D) 64 years
 
Answer & Explanation Answer: A) 84 years

Explanation:

Let the age of ruler is x so that of son = x/2 (given)
Now according to the given condition
(x/6) + (x/12) + (x/7) + 5 + (x/2) + 4 = x
=> x = 84

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Filed Under: Problems on Ages
Exam Prep: AIEEE , Bank Exams , CAT , GATE , GRE
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6 8307
Q:

If Hema walks at 12 km/hr instead of 8 km/hr, she would have walked 20 km more. The actual distance travelled by Hema is ?

A) 40 kms B) 30 kms
C) 46 kms D) 32 kms
 
Answer & Explanation Answer: A) 40 kms

Explanation:

Let the actual distance travelled be x km.
Then x/8=(x+20)/12
=> 12x = 8x + 160
=> 4x = 160
=> x = 40 km.

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Filed Under: Time and Distance
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5 8305
Q:

If a parallelogram with area p, a triangle with area R and a triangle with area T are all constructed on the same base and all have the same altitude, then

A) P=R B) P=A
C) P=A/2 D) P=2R
 
Answer & Explanation Answer: A) P=R

Explanation:

let each have base = b and height = h
then p = b*h, R = b*h and T = (1/2) * b*h
so P = R, P = 2T and T = (1/2)*R are all correct statements

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Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

2 8301
Q:

A bill falls due in 1 year.The creditor agrees to accept immediate payment of the half and to defer the payment of the other half for 2 years.By this arrangement ins Rs.40.what is the amount of the bill,if he money be worth 12.5%?

A) 1200 B) 3000
C) 3600 D) 1300
 
Answer & Explanation Answer: C) 3600

Explanation:

Let the sum be Rs. x. Then,   

x2+x2×100100+252×2-x×100100+252×1 = 40  

(x/2)+(2x/5)-(8x/9) = 40  

x=3600.  

Therfore, amount of bill=Rs. 3600.

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Filed Under: True Discount

5 8298
Q:

A person got Rs.48 more when he invested a certain sum at compound interest instead of simple interest for two years at 8% p.a. Find the sum  ?

A) Rs.8000 B) Rs.6500
C) Rs.7500 D) Rs.5000
 
Answer & Explanation Answer: C) Rs.7500

Explanation:

p=d×1002R2 

Where d is difference, r is rate of interest. 

=> (48 x 100 x 100) / 8 x 8 = Rs.7500

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Filed Under: Simple Interest
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6 8295
Q:

A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained?

A) 1/5 B) 1/6
C) 1/7 D) 1/8
 
Answer & Explanation Answer: B) 1/6

Explanation:

Total work is given by L.C.M of 72, 48, 36

Total work = 144 units

Efficieny of A = 144/72 = 2 units/day

Efficieny of B = 144/48 = 3 units/day

Efficieny of C = 144/36 = 4 units/day

 

According to the given data,

2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100

3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54

p = 198/16.5

p = 12 days.

 

Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day

(C+D) in p days = (4 + 6) x 12 = 120 unit

Remained part of work = (144-120)/144 = 1/6.

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