MATHS : QUIZ for SSC

1.The banker’s discount on Rs. 1600 at 15% per annum is the same as true discount on Rs. 1680 for the same time and at the same rate. The time is:
A. 3 months
B. 4 months
C. 6 months
D. 8 months

2.The banker’s discount on a sum of money for 1 years is Rs. 558 and the true discount on the same sum for 2 years is Rs. 600. The rate percent is:
A. 10%
B. 13%
C. 12%
D. 15%

3. The present worth of a sum due sometimes hence is Rs.5760 and the baker’s gain is Rs.10. What is the true discount?
A. Rs. 480
B. Rs. 420
C. Rs. 120
D. Rs. 240
4. A bill for Rs. 3000 is drawn on 14th July at 5 months. It is discounted on 5th October at 10%. What is the Banker’s Discount?
A. Rs. 60
B. Rs. 82
C. Rs. 90
D. Rs. 120
5.Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
A. Rs. 6400
B. Rs. 7200
C. Rs. 6500
D. Rs. 7500
6.If simple interest on a certain sum of money for 8 years at 4% per annum is same as the simple interest on Rs. 560 for 8 years at the rate of 12% per annum then the sum of money is
A. Rs.1820
B. Rs.1040
C. Rs.1120
D. Rs.1680
7.The compound interest on a sum for 2 years is Rs. 832 and the simple interest on the same sum for the same period is Rs. 800. The difference between the compound and simple interest for 3 years will be
A. Rs. 48
B. Rs. 66.56
C. None of these
D. Rs. 98.56
8.John invested money in two schemes A and B offering compound interest @ 5 p.c.p.a. and 10 p.c.p.a. respectively. If the total amount of interest accrued through two schemes together in two years was Rs. 2075 and the total amount invested was Rs. 15,000, find out the amount invested in Scheme A?
A. Rs. 10000
B. Rs. 8000
C. Rs. 12000
D. Rs. 14000
9. If the cost of x metres of wire is d rupees, then what is the cost of y metres of wire at the same rate?
A. Rs. (xd/y)
B. Rs. xd
C. Rs. (yd/x)
D. Rs. yd
10.A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. If the smaller wheel has made 21 revolutions, what will be the number of revolutions made by the larger wheel?
A. 15
B. 12
C. 21
D. 9

ANSWERS WITH SOLUTION:
1(B)Explanation:
S.I. on Rs. 1600 = T.D. on Rs. 1680.
 Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is on Rs. 1600 at 15%.
 Time = ((100 x 80)/(1600 x 15)year = 1/3 year = 4 months.

2( C)Explanation:
B.D. for  3/2 years =  Rs. 558.
B.D. for 2 years = (558 x 2/3 x 2)
                                      = Rs. 744
T.D. for 2 years = Rs. 600.
 Sum = (B.D. x T.D)/(B.D. – T.D) = Rs. (744 x 600/144) = Rs. 3100.
Thus, Rs. 744 is S.I. on Rs. 3100 for 2 years.
Rate = (100 x 744)/(3100 x 2)%= 12%

3(D)

4( A)Explanation :F = Rs. 3000
R = 10%
Date on which the bill is drawn = 14th July at 5 months
Nominally Due Date = 14th December
Legally Due Date = 14th December + 3 days = 17th December
Date on which the bill is discounted = 5th October
Unexpired Time
= [6th to 31st of October] + [30 Days in November] + [1st to 17th of December]
= 26 + 30 + 17 = 73 Days
=73/365 year = 1/5 year
BD = Simple Interest on the face value of the bill for unexpired time =FTR/100
=3000×1/5×10/100 =30×1/5×10 = Rs. 60

5( A)Explanation :
Let the investment in scheme A be Rs.x
and the investment in scheme B be Rs.(13900 – x)
We know that SI = PRT/100
Simple Interest for Rs.x in 2 years at 14% p.a. = x×14×2/100=28x/100
Simple Interest for Rs.(13900 – x) in 2 years at 11% p.a
. =(13900-x)×11×2/100=22(13900-x)/100
Total interest = Rs.3508
28x/100+22(13900-x)/100=3508
28x+305800-22x=350800
6x=45000
x=450006=7500
Investment in scheme B = 13900 – 7500 = Rs.6400

6(D)Explanation :
Let the sum of money be x
then
x×4×8/100=560×12×8/100
x×4×8=560×12×8
x×4=560×12
x=560×3=1680

7(D)Explanation :
Given that simple interest for 2 years is Rs.800
i.e., Simple interest for 1st year is Rs.400
and simple interest for 2nd year is also Rs.400
Compound interest for 1st year will be 400
and Compound interest for 2nd year will be 832 – 400 = 432
you can see that compound interest for 2nd year is more than simple interest for 2nd year by 432 – 400 = Rs.32
i.e, Rs. 32 is the interest obtained for Rs.400 for 1 year
Rate, R = 100×SI/PT= (100×32)/(400×1)= 8%
Difference between compound and simple interest for the 3rd year
= Simple Interest obtained for Rs.832
=PRT/100=832×8×1/100= Rs. 66.56
Total difference between the compound and simple interest for 3 years
= 32 + 66.56 = Rs.98.56

8(A)Explanation :
Let the investment in scheme A be Rs.x
Then the investment in scheme = Rs.(15000-x)
Compound interest for Rs.x = P(1+R/100)^T-P= x(1+5/100)^2-x
Compound interest for Rs. (15000-x) = P(1+R/100)^T-P=
(15000-x)(1+10/100)^2-(15000-x)
Total compound interest = x(1+5/100)^2-x+(15000-x)(1+10/100)^2-(15000-x)
=x(1+10/100+25/10000)-x+(15000-x)(1+20/100+100/10000)-(15000-x)
=x(10/100+25/10000)+(15000-x)(20/100+100/10000)
=1025x/10000+(15000-x)((2100/10000)
=1025x/10000+15000×2100/10000-2100x/10000
=-1075x/10000+3150
Given that total compound interest = Rs. 2075
-1075x/10000+3150=2075
=>1075x/10000=1075
=>x/10000=1
=>x=10000
i.e, Amount invested in Scheme A = Rs. 10000

9(C)Explanation :
cost of x metres of wire = Rs. d
cost of 1 metre of wire = Rs.(d/x)
cost of y metre of wire = Rs.(y×d/x)=Rs. (yd/x)

10(D)Explanation :
Let the number of revolutions made by the larger wheel be x
More cogs, less revolutions (Indirect proportion)
Hence we can write as
Cogs 6 : 14} :: x : 21
 6×21= 14×x
6×3 = 2×x
3×3 = x
x=9

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48 Comments

  1. Mam…I secured 305 marks (in both Tier 1 and Tier 2)….Can I be called for SSC CGL Interview ??? Please mam…enquire and reply me

  2. Pls solve this problem
    A cylindrical vessel of diameter 12 cm contains some water . if two spheres of radii 6 cm each are lowered into the water until they are completely immersed then the water level in the vessel will rise by?
    A. 4
    B. 9
    C. 12
    D.6
    Please provide me the solution for this problem .

  3. I don't know exactly , I think 4 is given in some of the sites, what is the formula

  4. let rise of water b H

    r1 = radius of cylinder=12/2=6 cm

    r2=radius of sphere= 6 cm

    then vol of cylinder=2*vol of sphere

    pie*(r1^2)*H =2*(4/3)pie*(r2)^3

    (r1^2)*H =2*(4/3)*(r2)^3

    6*6 *H= 2*(4/3)*6*6*6
    H=2*(4/3)*6
    H=16 cm
    if u take radius of cylinder 12 cm
    then H=4 cm
    may b in Q in place of diameter, there is radius

  5. Just now I got the answer
    For this formula is
    Rice in water level = volume of sphere/area of cylinder
    =2*4/3*pi*6*6*6/(pi*12*12).

  6. This question is given in SSC cgl tire2 2015 , same model given in DC cgl 2012 paper, if u have kiran model papers check it , u will get good idea

  7. Hw i can guide u ..???
    I gave last year'Sbi po before that i applied but could nt gave bcz exam centre was vry distant.

  8. but u had given last yr na…..
    it means u hv some knowledge abt sbi po….. nd i want that knowledge

  9. okk…don't b sad yaar u r a laborious girl. u can clr it
    now wt is yr strategy for this year sbi po ?

  10. Nt this time will plan aftr giving nicl asst. Nd fci
    I hv to prep hard for reas
    Nd di only .remaining section i will clr easily..
    Meti tarif kam kiya kro…. :))

  11. Gussa to aata h mje bt i nvr showed it on SA except one single time on a boy other then u
    I nvr been angry with u… i knw

  12. It's based on archimedis principle. Simply it's answer is..

    => Rise of level= (total volume putting inside)/(area of base of the water vessel used)

    => 2(4/3 pi r^3) / pi r^2

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