36 HOUR PLACEMENT CLASS SHEET
QUANTITATIVE APPTITUDE
1
Simple Interest & compound intrest,
Algebra, percentage, profit & loss
1.
A sum was put at a certain rate for 2 years.
Had it been put at 6% higher rate, it would
have fetched Rs.600 more. Find the sum
a) 10,000
b) 2500
c) 5000
d) 2900
2.
Interest on a certain sum at a certain rate is
4/25 of the sum. If the number representing
rate percent and time in years be equal, then
the rate is
1
a) 6%
b) 7 %
2
c) 4%
d) 8%
3.
4.
5.
6.
7.
8.
If the difference between CI and SI on a
certain amount of money is Rs 256 at 16% pa
for 2 years .Find the amount?
a) 10,000
b) 12,000
c) 8500
d) 9000
Mr. Senthil borrowed Rs 16,000 at 4% pa on
SI and immediately he rent the whole sum to
Mr. Bala at 4% pa on CI. What will be the
gain of Mr. Senthil after 4 years?
a) 158 Rs
b) 30 Rs
c) 300 Rs
d) 3000 Rs
Find the ratio earned by simple interest for a
certain sum in certain rate for 4 years and 3
years?
a) 4:3
b) 3:4
c) 1:2
d) 2:3
A man invested 2/3 of his capital at 6%, ¼ at
8% and the remainder at 9%. If his annual
income is Rs 324. The capital is
a) 3600
b) 2100
c) 2400
d) 4800
Find the time for which Rs 10000 amounts to
Rs 14641 at 10% pa compounded interest?
a) 1
b) 2
c) 4
d) 8
At what rate percent per annum, CI will Rs
2500 amount to Rs 6400 in 2 years
a) 20 %
b) 40 %
c) 60 %
d) 25 %
9.
Kesavan borrowed a certain sum at the rate of
15% pa. If he paid at the end of 3 years Rs
1230 as Interest Compounded annually find
the sum
a) 2476 Rs
b) 3500 Rs
c) 3200 Rs
d) 2365 Rs
10. A sum of money compounded annually
amounts to Rs 7290 in 3 years and Rs 17280
in 6 years. Find the rate of interest?
a) 3.33%
b) 33.3%
c) 333.3%
d) 3%
3
2
11. Evaluate x  6.5 x  13.5 x  9
x
a)
c)
2
 17.5 x  24 
𝑥−3
b)
16
(𝑥−2)𝑥
16
d) x-1.5
𝑥−1
3
(𝑥−3)(𝑥−2)
4
12. Evaluate √−10648 + √4096
a) 24
b) -36
c) -14
d) 54
13. If there is a rectangular window whose length
is 4 times the width and the sum of measures
of all sides of the window is 600 cm. Find the
length of the window
a) 260
b) 320
c) 340
d) 240
14. A pen and a refill together costs Rs 7.50
paise. If the pen costs Rs 3 more than the
refill find the price of the pen.
a) 5.25
b) 6.00
c) 5.75
d) 5.00
15. Sum of two numbers is 81, difference
between the two number is 31. Find the
largest of the two numbers
a) 55
b) 26
c) 56
d) 25
16. Find the value of P, for which the roots of the
quadratic equation x2+24x+P=0, are equal
and real?
a) 256
b) 576
c) 144
d) 72
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36 HOUR PLACEMENT CLASS SHEET
17. (2P)4. P2 = 1024 find P
a) 64
c) 8
b) 512
d) 2
18. If p+q+r = -6, pq+qr+rp=4, find p2+q2+r2
a) 19
b) 18
c) 28
d) 24
19. If 2x+3y=8 and xy=4, find the value of
4x2+9y2.
a) 20
b) 16
c) 15
d) 10
20. For which value of K does the following pair
of equators yield a unique solution of x such
that the solution is positive
x2-y2 = 0
(x-k)2+y2 = 1
a) 2
b) 0
b) √2
d) −√2
21. One of the +ve roots of a quadratic equation
ax2+mx+c = 0 where a=1, is twice that of the
other. The sum of the co-efficient of x and
constant term is 2. What is the value of m in
the eqn?
a) -4
b) 4
c) -6
d) 6
22. The number of real solutions of the equation
6−𝑥
𝑥
=2+
is
𝑥 2 −4
𝑥+2
a) two
b) one
c) zero
d) None of these
1
2
QUANTITATIVE APPTITUDE
other a loss of 15%. How much % loss or %
gain made in the whole transaction?
a) 2.25% loss
b) 2.25% gain
c) 8% loss
d) 8% gain
27. A person sells 36 oranges per rupee and
suffers a loss of 4%. Find how many oranges
per rupee is to be sold to have gain of 8%
a) 28
b) 32
c) 62
d) 50
28. A shopkeeper purchases 10 kg of rice at Rs
600 and sells at a loss as much the selling
price of 2 kg of rice. Find the sale rate of rice.
a) 56 Rs/Kg
b) 30 Rs/Kg
c) 40 Rs/Kg
d) 50 Rs/Kg
29. A man sells a book at a profit of 20%. If he
had bought it at 20% less and sold if for Rs
18 less, he would have gained 25%. Find the
cost price of the book?
a) 45 Rs
b) 200 Rs
c) 90 Rs
d) 250 Rs
30. A shop keeper earns a profit of 12% on
selling a book at 10% discount on the printed
price. The ratio of the CP and the printed
price of the book is
a) 45:56
b) 50:61
c) 99:125
d) 36:79
31. What is the equivalent discount of the 20%,
10%, 5% discount series
1
a) 35
b) 17
2
c) 28
d) 31.6
1
23. If z+ = 1, z64 + 64 is equal to
𝑧
𝑧
a) 0
b) 1
c) -1
d) -2
24. The Polynomial (ax2+bx+c) (ax2-dx-c).ac ≠ 0
has
a) four real zeros
b) Atleast 2 zeros
c) Atmost 2 zeros
d) No real zeros
25. The condition that both the roots of the
equations ax2+bx+c = 0 are zero is
a) b≠0
b) c≠0
c) Either b or c (not both )=0
d) b=c=0
26. Two watches were sold at Rs.850 each. On
one a the again of 15% is made and on the
32. An article when sold for Rs 200 fetches 25%
profit. What could be the percentage
profit/loss if 6 such articles are sold for Rs
1056?
a) 10% loss
b) 10% profit
c) 5% loss
d) 5% profit
33. A milkman makes a profit of 20 % on the sale
of milk. If he were to add 10% water to the
milk by what % would his profit increase.
a) 22
b) 32
c) 33
d) 10
34. The percentage profit earned by selling an
article for Rs 2480 is equal to the percentage
loss incurred by selling the same article for
Rs 1980. At what price should the article be
sold to make 20 % profit?
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36 HOUR PLACEMENT CLASS SHEET
a) Rs 3800
c) Rs 2780
3
QUANTITATIVE APPTITUDE
b) Rs 3456
d) Rs 2676
35. If the cost price of 15 balls
selling price of 12 balls,
percent?
a) 50%
b)
c) 75%
d)
is equal to the
Find the gain
25%
30%
36. A man buys 200 oranges for Rs 10. How
many oranges for a rupee can he sell so that
his profit percentage is 25%?
a) 10
b) 16
c) 15
d) 20
37. If the population of town is decrease by 10%
and 20% in 2 successive years. What is the
net % increase or decrease in the population?
a) 26% increase
b) 32% decrease
c) 30% increase
d) 28% decrease
1
38. A man loses 12 % of his money and after
2
spending 70% of reminder, he is left with Rs
210. How much did he have with him at first.
a) 350
b) 342.85
c) 340.75
d) 200
39. Rahu’s salary is decreased from Rs 700 to Rs
630. Find decrease percentage?
a) 20%
b) 25%
c) 10%
d) 5%
40. By selling an article for Rs 360, the loss
incurred is 10%. At what minimum price
should he sell that article to avoid loss
a) 400 Rs
b) 500 Rs
c) 600 Rs
d) 480Rs
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4
36 HOUR PLACEMENT CLASS SHEET
QUANTITATIVE APPTITUDE
SOLUTION
1.
2.
Ans: c
Solution:
𝑃×𝑟×2
𝑃×(𝑟+6)×2
I1 =
,
I2 =
100
100
I2 - I1 = 600
𝑃×(𝑟+6)×2 𝑃×𝑟×2
= 600 ; 12P = 60,000
100
100
P = 5000
7.
Ans: c
Solution:
A = P (1 +
=
P
4
=
8.
P
25
2
𝑟=
10000
Ans: a
Solution:
CI – SI =
256 =
× 10
5
20
16
1+
𝑟
(
𝑃𝑅
100
9.
× 100
5
800
Ans: d
Solution:
1230 = P
100
:
– 1]
) – 1]
1230 =
1000000
= P(0.5208)
P = 2365
4
10. Ans: b
Solution:
7290 =P(1 +
1728
100
)n – P
100
1520875−1000000
× 4 = 2560
I1 : I 2
𝑃×3×𝑅
𝑟
100
15
[(1+ )3
100
115 3
1230 = P [(
4
Ans: b
Solution :
729
12
𝑃×4×𝑅
= (1 +
( ) = 1+
100
4
3:4
3
= 1+
𝑟
100
9
𝑟
𝑟
)3 14580=P(1+
100
𝑟 3
𝑟
100
)
100
100
400
100+r =
3
400
r=
- 100
3
r = 33.3%
6
100
+
+
2𝑥
1200
81𝑥
𝑥
4
+
300
100
48𝑥+24𝑥+9𝑥
1200
8
CI = P (1 +
CI = 16000 {[1+ ]4 – 1}
100
CI = 2717.73
CI – SI = 157.7
 158
12𝑥
5
)
Ans: a
Solution:
SI = 16000 ×
3
=
)2
r=
- 100
5
r = 60%
)
100 100
𝑃×16
(
𝑟
100
8
100 + 𝑟 =
𝑅
)n
100
10 n
)
100
100
P=
16×16
P = 10,000
Ans: d
Solution:
2𝑥
×
𝑟
= P (1 +
6400 = 2500(1 +
100 100
256×100×100
6.
)n
∴n=4
Ans: c
Solution:
𝑟
𝐴 = 𝑃(1 + )n
=>
5
𝑟 = 4%
5.
14641
𝑟
100
= 4800
11
11
( )4 = ( )n P
10
10
4
100
4.
81
= P (1 +
Ans: c
Solution:
SI =
25
𝑃×𝑟×𝑟
3.
324 ×1200
𝑥=
= 324
×
8
100
9𝑥
1200
= 324
+
1
12
= 324
×
9𝑥
100
= 324
11. Ans: b
Solution:
𝑥 3 −6.5𝑥 2 +13.5x−9
𝑥 2 −17.5𝑥+24
=
=
(𝑥−3)(𝑥−1.5)(𝑥−2)
(𝑥−1.5)(𝑥−16)
(𝑥−3)(𝑥−2)
𝑥−16
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)6
36 HOUR PLACEMENT CLASS SHEET
12. Ans: c
Solution:
3
√−10648 = (-10648) 1/3 = -22
= (4096)1/4 = 8
√4096
3
4
√−10648 + √4096 = -22 +8 = -14
4
QUANTITATIVE APPTITUDE
5
19. Ans: b
Solution:
(2x + 3y)2 = 4x2 + 9y2 + 12xy
64 -48 = 4x2 + 9y2
4x2 + 9y2 = 16
13. Ans: d
Solution:
Length = 4x breadth =x
2(x+4x) = 600
10x=600
x=60
Length = 240 cm
20. Ans: c
Solutions:
x2 = y2
2x2 – 2kx +k2 – 1 = 0
going by choices
Put x = √2 ;
k2 - 2 √2 k + 3 = 0 ie K = √2
14. Ans: a
Solution:
Let the price of refill = x
x+x+3 = 7.50
2x+3 = 7.50
pen = 5.25
2x = 4.5
x = 2.25
21. Ans: c
Solution:
Let the roots be ∝ , 2 ∝ where ∝ > 0
∝ +2 ∝ = −𝑚 => 𝑚 = −3 ∝
2 ∝2 = C
Since m + c =2 => 2 ∝2 - 3∝ = 2
1
∝= − ,2
2
∝>0
∴∝= 2
∴ 𝑚 = −3 ∝ = −6
∝= 2
15. Ans: c
Solution:
Let no’s = x-y
x + y = 81
x – y = 31
22. Ans: b
Solution:
2x = 112
x = 56
y + 56 = 81
y = 81-56
y = 25
6  x 2  x  2  x

x2  4
x2
16. Ans: c
Solution:
Roots are real and equal
b2 - 4ac = 0
ie 242 – 4(ac) = 0
2
24 – 4 × 1 × 𝑃 = 0
576 – 4P = 0
P = 144
17. Ans: d
Solution:
16.P6 = 1024
1024
P6 =
= 64
16
P=2
18. Ans: c
Solution:
p2 + q2 + r2 = 36 – 4(2)
= 28
with x+2 ≠ 0 and x – 2 ≠ 0
6−𝑥
=>
= 3𝑥 + 4.
𝑥−2
6-x = (3x+4)(x-2)
3x2 – x – 14 = 0
x = -2 x = 7/3
But (x+2) ≠ 0 hence, only one root is
possible
23. Ans: c
Solution:
1
Z+ =1
Z2 – Z + 1 = 0
𝑧
(Z+1) (Z2 – z + 1) = 0 if Z ≠ −1
Z3 + 1 = 0
Z3 = -1
Z ≠ −1
Z 64  1
Z 64
 Z 63 .Z 
1
Z .Z
63
  Z 3  .Z 
21
 Z  1
1
Z 
3 21
.Z
Z
 1
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36 HOUR PLACEMENT CLASS SHEET
24. Ans: b
Solution:
(ax2+bx+c)( (ax2-dx-c) = 0
∴ either ax2+bx+c = 0 or ax2-dx-c = 0 or
both.
∴ roots of ax2+bx+c = 0 will be real if
b2-4ac>0. Similarly for ax2-dx-c roots
will be real if d2+4ac>0. Now atleast one
of the two condition will hold true since
either 4ac will be greater than zero or
less than or equal to zero.
∴ Atleast 2 zeros will be there
25. Ans: d
Solution:
ax2+bx+c = 0
for both roots to be zero.
sum of roots = -b/a = 0
∴𝑏=0
𝑐
=0
𝑎
10
1
36
Using formula
=
1/36
100+(−4)
1
S2 =
=
𝑆1
100+𝑥1
𝑆2
=
112
1125
14
31. Ans: d
Solution:
20%
10%
1125
14
: 100 => 45:56
15%
80%
90%
95%
 0.80 × 0.90 × 0.95
 0.6840
 1 – 0.6840
 0.3160
 31.6
= -( )2
10
=
- 2.25%
= loss = 2.25%
27. Ans: b
Solution:
Unit price =
30. Ans: a
Solution:
Let the printed price of the book be Rs
100. After the discount of 10%
SP = Rs 90
Profit earned = 12%
100
CP =
× 90
cp: printed price =
26. Ans: a
Solution:
𝑥
Overall % loss = -( )2
15
Rs 80 Sells 80 × 1.25
= 100
at 25% profit
S1 – S2 => Rs 20 when CP is 100
S1 – S2 => Rs 18, so the CP is 90
∴ CP = 90

𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠 =
∴𝑐=0
QUANTITATIVE APPTITUDE
6
𝑆2
100+𝑥2
100+8
32. Ans: b
Solution:
200
CP of 1 article =
× 100
125
= Rs 160
CP of 6 article = 6 × 160
= 960
Profit = 1056 – 960 => 96
96
Profit % =
× 100 = 10%
960
32
33. Ans: b
Solution:
200
a+ b +ab/100 =>20 + 10 +
100
= 32%
28. Ans: d
Solution:
Loss = CP – SP
Let S.P = x
2 × x = 600 – 10 × x
x = 50 Rs/Kg
29. Ans: c
Solution:
Assume CP of book = 100
100 Sells 100 × 1.2 = 120 (S1)
at 20 % profit
If he bought 20 % less
34. Ans: d
Solution:
Let CP be x
Then (2480 - x)100/x = (x - 1980)100/x
2480 – x = x -1980
2x = 4460
x = 2230
S.P = 120% of 2230
=120 × 2230/100 =>2676
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36 HOUR PLACEMENT CLASS SHEET
7
QUANTITATIVE APPTITUDE
35. Ans: b
Solution:
Let C.P of 12 balls Re 1.
C.P of 12 balls = Rs 12. S.P
of 12 balls
= Rs 15
100
=3×
12
= 25%
36. Ans: b
Solution:
A buys 200 oranges
CP of 200 oranges = 10 Rs
For 1Rs he gets 20 oranges
(200/10) SP of 200 oranges to
get profit 25% = 10 × 1.25
= 1.25 rs
So of 1 rs he sells 16 oranges
(200/12.5)
= 16 oranges
37. Ans: d
Solution:
x = -10
x = -20
𝑥𝑦
The net % change = (x + y + ) %
100
200
= (-10 -20 + )%
100
= -28
= 28 % decrease
38. Ans: c
Solution:
1
A man loses 12
2
Let man has x Rs
Money left after losing 12.5%
= 0.875 × 𝑥
Money left after spending 70% of
remaining money = 0.7 × 0.875𝑥
= 210
x = 342.85 Rs
39. Ans: c
Soluiton:
[(700 – 630)/700] × 100
= 10%
40. Ans: a
Solution:
360
Price to avoind loss x =
0.9
= 400
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36 HOUR PLACEMENT CLASS SHEET
QUANTITATIVE APPTITUDE
8
ANSWER KEY
1)
c
2)
c
3)
a
4)
a
5)
b
6)
d
7)
c
8)
c
9)
D
10)
b
11)
b
12)
c
13)
d
14)
a
15)
c
16)
c
17)
d
18)
c
19)
B
20)
c
21)
c
22)
b
23)
c
24)
b
25)
d
26)
a
27)
b
28)
d
29)
C
30)
a
31)
d
32)
b
33)
b
34)
d
35)
b
36)
b
37)
d
38)
c
39)
C
40)
a
HO: No 15, 1st floor, Green Palace Building, South Dhandapandistreet, T.Nagar, Chennai – 600017.
Email: [email protected] website: www.axiomacademy.com