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MATHS ­­ PRACTICE SET 11 ­­ SOLVED (Combined
Graduate Level Exam (CGLE)
4/14/2016
MAXIMUM TIME: 20 minutes
O
M
1. What is the unit place of 122173?
.C
A) 6
B) 3
C) 2
D) 1
M
2. Find the sum of first 50 terms in the series: 5 + 3 ­ 7 + 5 + 3 ...
R
U
A) 1
B) 8
C) 24
D) 40
FO
3. The greatest common divisor of (33)333 + 1 and (33)334 + 1
c) (33)333 + 1
d) 20
AM
a) 2
b) 1
SS
C
A) 0
B) 1
C) 2
D) 3
EX
4. If x + 1/x = √3; then the value of x18 + x12 + x6 + 1 is ?
5. If a/b = b/c = c/d; then (b³ + d³ + c³) /(a³ + b³ + c³) = ?
A) d/a
B) d/c
C) a/b
D) a/d
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6. In what proportion may these kinds of tea of prices @Rs 80, Rs 70 and Rs 50 per
kg be mixed to produce a mixture of Rs 60 per kg?
A) 2:2:3
B) 1:1:3
O
M
C) 1:1:2
D) 2:3:2
7. Out of 10 teachers of a school one teacher retires and in his place a new teacher
of age 25 years joins. AS a result average age of teachers is reduced by 3 years. The
.C
age in years of the retired teacher is?
M
A) 60 years
B) 62 years
D) 55 years
FO
8. If (a+b) : (b+c) : (c+a) = 6 : 7 : 8
and a+b+c = 14 then, the value of 'c' ?
R
U
C) 58 years
A) 6
B) 7
C) 8
AM
D) 9
9. A reduction of 20% in the price of sugar enables a purchaser to obtain 3 kg more
EX
for 120; the original price of sugar is?
A) Rs 8 per kg
C
B) Rs 7 per kg
C) Rs 10 per kg
D) Rs 12 per kg
SS
10. An article is sold at a certain fixed price. By selling it at 2/3 of that price, one
loses 10%. The gain percent on selling it at the original price is?
A) 25%
B) 15%
C) 28%
D) 35%
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11. Rs.3000 amounts to Rs.6750 after a certain period of time when the interest is
applied annually. What will be the amount if the time period is taken half of the
previous time?
O
M
A) Rs 3750
B) Rs 4000
C) Rs 4550
D) Rs 4875
12. Ram can do a work in 5 days less than the time taken by Shyam to do it. If both
.C
of them together take 100/9 days, then the time taken by Shyam alone to do the
same work (in days) is?
M
A) 15
R
U
B) 25
C) 20
D) 30
13. Train A leaves Meerut at 4 pm and arrives Ghaziabad at 5pm. Train B leaves
FO
Ghaziabad at 4 pm and arrives Meerut at 5.30 pm. At what time both the trains
would meet each other?
B) 4:36 pm
C) 4:25 pm
D) 4:50 pm
AM
A) 4:00 pm
EX
14. A solid right circular cylinder and a solid hemisphere stand on equal bases and
have the same height. The ratio of their whole surface area is?
C
A) 4:3
B) 3:4
C) 2:3
SS
D) 3:2
15. A cylindrical can whose base is horizontal and is of internal radius 3.5 cm,
contains sufficient water so that when a solid sphere is placed inside, water just
covers the sphere. The sphere fits in the can exactly. The depth of water in the can
before the sphere was put is
A) 7/3 cm
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B) 14/3 cm
C) 35/3 cm
D) 17/3 cm
O
M
16. In a 8 cm equilateral triangle two other equilateral triangles are made from its
centroid so that they divide the main equilateral triangle in three equal areas;
then find the length of side of the smaller triangle
A) 4
B) 2√2
.C
C) 8/√3
D) 8/3
R
U
M
17. What is the difference between the areas of the circumcircle and incircle of a
regular polygon of length '2a' as its side.
FO
A) πa²
B) a²
C) 2a²
D) 4a²
D) 7.2
EX
A) 6.5
B) 7.6
C) 8
AM
18. Radius of a circle is 10. Two chords of 16 cm and 17 cm cut each other
perpendicularly; then distance between center of the circle and the intersecting
point of two chords is ?
C
19. sin+ cosθ = √2 cosθ. Find value of cosθ ­ sinθ
SS
A) √3 cosθ
B) √3 sinθ
C) √2 cosθ
D) √2 sinθ
20. sin20sin40sin60sin80 = ?
A) 1/16
B) 1/8
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C) 3/16
D) 1/2
1. It's easy to get that 122173 will give the same unit place as 2173 will.
O
M
SOLUTION
Now 2173 = (2172)*2 = [(24)43]*2 (as 172 is fully divisible by 4; we took it because 2 raised to
.C
power of greater than 4's multiples starts giving repeated unit places.
M
You'll see 24 gives 6 as unit place, so (24)43 will also give 6 as unit digit. You see it's multiplied
by 2.
R
U
So the unit digit that required = unit digit in 6*2 i.e. 2 (option 'C')
FO
2. The first three terms i.e. (5 + 3 ­ 7) are making one set of the series
Let's see how many such sets are there in the series
AM
Number of terms given = 50
So, the number of sets = 50/3 = 16 +2, means 16 is the number of full sets of three terms each
and 2 being the first two terms.
EX
Sum of one set = 5 + 3 ­ 7 = 1
Sum of the next 2 terms = 5 + 3 = 8
Hence the sum of all 50 terms = Sum of all full sets + Sum of the remaining 2 terms
C
= 16*1 + 8 = 24 (option 'C')
SS
3. The first term of the second expression (33)334 can be re­written as
3^[(3^333)*(3^1)]
Now let 3333 be y
Therefore the given expressions are 3y + 1 and 33y + 1
=> 3y + 1 and (3y)3 + 1
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If seen carefully they are in the form of x+1 and xn +1
We know an expression in the form of xn +1 is divisible by an expression of the form x +1 for all
odd values of 'n'
O
M
Here n = 3; which is odd.
So the second expression is divisible by the first; and by look we can see it's the greatest of all
the given options
.C
Hence (33)333 + 1 (option 'C') is correct.
x4 + 1 = x2....................(1)
R
U
4. x + 1/x = √3
squaring both sides and taking LCM we get:
M
NOTE: We don't need to check other options as it's the greatest of all.
FO
multiply both sides by x² (We have to multiply it by x2 so that the degree of the equation is same
as the degree of the last term in x in the expression is, i.e. 6)
=> x6 + x² = x4
=> x6 = x4 ­ x²
AM
So the equation will be (x4*x2) + (1*x²) = (x²*x²)
EX
Now the given expression can be re­written as:
(x6)3 + (x6)2 + x6 + 1
C
= (x4 ­ x2)3 + (x4 ­ x2)2 + (x4 ­ x2) + 1 (by putting the value of x6)
SS
from (1) we get x4 ­ x2 = ­1
put it in the above and solve, answer = 0 (option 'A')
5. Let a/b = b/c = c/d = k
=> c = dk, b = dk², a = dk³
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Now the given expression
= (b³ + d³ + c³) /(a³ + b³ + c³)
= [(d3*k6) + (d3) + (d3*k3)]/[(d3*k9) + (d3*k6) + (d3*k3)]
O
M
= [d3(k6 + 1+ k3)]/[d3(k9 + k6 + k3)]
= (k6 + 1+ k3)/(k9 + k6 + k3)
.C
= (k6 + 1+ k3)/[k3(k6 + k3 + 1)
= 1/k3
M
= (b/a)(c/b)(d/c)
R
U
= d/a (option 'A')
FO
6. We see here that we have three quantities having different values. When the case is like that
take any two quantities (say a and b) and then find the ratio using allegation method or the way
you find easier. Similarly take the third quantity and any of a and b, and find their ratio as you
did earlier.
AM
Now is the turn of finding the combined ratio of those two ratios calculated above.
Now see how it will work when the above question is considered.
EX
a c b c
80 50 70 50
60 60
C
10 20 10 10
Therefore the combined ratio of these two ratios = 10 : 10 ; (20+10) = 10 : 10 : 30 = 1 : 1 : 3
SS
(option 'B')
You'll see whatever proportion of above two mixtures we take, the price of the resulting mixture
will always be same as both the mixtures cost Rs 60 per kg.
7. Age of the retired teacher = Age of the new teacher + Total decrease in age = 25 + (10*3) = 55
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years (option 'D')
8. (a+b) : (b+c) : (c+a) = 6:7:8 (given)
O
M
Now let
a+b = 6x .........(i)
b+c= 7x ..........(ii)
.C
c+a = 8x .........(iii)
Now adding all these three equations, (a+b) + (b+c) + (c+a) = 6x + 7x + 8x
=> 2(a+b+c) = 21x
R
U
M
=> 2*14 = 21x ­­­(given, a+b+c=14)
=> x= 28/21 = 4/3
But given, a+b+c = 14
=> (a+b) + c= 14
=> 8 + c = 14
AM
Hence c= 14 ­ 8 = 6 (option 'A')
FO
From (i)
a+b = 6x = 6*(4/3) = 8
EX
9. Reduction in the price = 20%
Therefore the purchaser's saving on = 20% of 120 = Rs 24
Thus he'll be able to buy 3 kg of sugar in 24 now
Hence the reduced price of Sugar = 24/3 = Rs 8 per kg
SS
C
According to 20% reduction in the price
If Rs 80 per kg is the reduced price the original price = 100 per kg
If 8 is the reduced price the original = (100/80)*8 = Rs 10 per kg (option 'C')
10. Let the cost price be Rs 100
Thus the selling price with 10% loss = Rs 90
Means this amount of Rs 90 is 2/3 of the original selling price
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So the the original selling price = 90 x 3/2 = Rs 135
Thus the gain = 135 ­ 100 = 35
O
M
But it's on the cost price of Rs 100
Hence the gain in percent = 35% (option 'D')
11. There is no mention of compound interest having been the basis. When such is the case it's
12. Let time taken by Shyam = x days
So time taken by Ram = x ­ 5 days
FO
Now Shyam's 1 day work = 1/x
And Ram's 1 day work = 1/(x ­ 5)
R
U
Interest for half the time = 3750/2 = 1875
Therefore the amount = 3000 + 1875 = 4875 (option 'D')
M
Interest for full time = 6750 ­ 3000 = 3750
.C
simple interest taken into consideration. If so the solution should be:
And their total work in 1 day = 9/100
EX
AM
Therefore 1/x + 1/(x ­ 5) = 9/100
Now instead of solving the equation algebraically try options in this equation as solving the
equation that way will be time consuming; of course the answer you will be getting is 25 (option
'B').
13. Let the distance be D km
SS
C
Therefore the speed of the train leaving Meerut = D/1 kmph (It takes 1 hour from Meerut to
Ghaziabad)
and the speed of the train leaving Ghaziabad = D/1.5 kmph (It takes 1.5 hour from Ghaziabad to
Meerut)
As they are moving in the opposite direction their relative speed = (D/1 + D/1.5) km/h
So the time taken in meeting each other = D/(D/1 + D/1.5) ­­­­ (Distance/Speed)
= 3/5 hour = (3/5)*60 minutes = 36 minutes
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Hence the time at which both the trains will meet each other = 4 pm + 36 minutes = 4:36 pm
(option 'B')
14. Whole surface area of the cylinder = 2πr(h+r)
O
M
Whole surface are of the hemisphere = 3πr²
Therefore, whole surface area of the cylinder : Whole surface are of the hemisphere = 2πr(h+r) :
.C
3πr²
But their height is the same and their bases being equal, so height of the cylinder and radius of
M
the hemisphere will be equal.
R
U
According to above, the ratio can be re­written as 2πr(r+r) : 3πr²
=> 4πr² : 3πr²
=> 4 : 3 (option 'A')
FO
15. According to the question
Volume of the can up to the length sphere is placed ­ Volume of the sphere = Volume of the can
up to the depth of water that was in the can before the sphere was placed inside
AM
πr²h ­ (4/3)πr³ =πr²H, where 'h' is the height of the can up to sphere is placed i.e. 2r = 2*3.5 =
7; r = 3.5 and H = depth of the water in the can before the sphere was put in (to find)
EX
Now dividing both sides of the above equation by πr²
h ­ (4/3)r = H
=> 7 ­ (4/3)*3.5 = H
=> 7/3 = H
SS
C
So 7/3 cm (option 'A') is the required answer.
16. area of the smaller triangle = (1/3)area of the bigger triangle
= ⅓[(√3/4)a²]; where 'a' is the side of the bigger triangle
= ⅓[(√3/4)*8²]
= (16/3)√3
Side of an equilateral triangle = √(4area/√3)
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Now letting the side of smaller triangle as 't'
t = √[4{(16/3)√3}/√3]
=>√[{(64/3)√3}/√3]
O
M
=> t = √64/3
=> 8/√3 (option 'C')
17. Let 'R' be the radius of circumcircle (line from centre to vertex) and 'r' be the radius of
incircle (perpendicular line from centre to the side)
M
'r' & 'a' as two of its sides.
Therefore according to Pythagoras, R² = r² + a²
.C
When you draw the figure, you'll, thus, find a right angle triangle with 'R' as the hypotenuse and
=> R² ­ r² = a²
AM
FO
Thus the difference in areas = πR² ­ πr²
=π(R² ­­ r²)
= πa² (option 'A')
R
U
As 'R' is the radius of the circumcircle and 'r' being the radius of the incircle, their areas
respectively = πR² and πr²
18. The line joining the centre of the cirlce to the intersecting point of two chords is actually the
EX
diagonal of the rectangle formed by the altitudes drawn from centre to the two chords
respectively.
C
We know that diagonal of a rectangle = √(length² + breadth²)
We see that the length and breadth of this rectangle actually are the altitudes drawn from centre
to the two chords respectively.
SS
So letting the altitude on the chord with 16 cm of length as H and the altitude on the chord with
17 cm of length as h we can write the above equation like this: diagonal = √(H² + h²)
But, H² = 10² ­ (16/2)² = 100 ­ 64 = 36
and h² = 10² ­ (17/2)² = 100 ­ 289/4 = 27.25
So according to the Pythagoras
the required difference = √(36 + 27.25) = √63.25 = 8 (appox)­­­­­­­­­­­­option 'C'
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19. sinθ + cosθ = √2cosθ
squaring
O
M
1 + sin2θ = 2cos²θ
= 2 ­ 2sin²θ
=2sin²θ = 1 ­ sin2θ
Thus, 2sin²θ = [cosθ ­ sinθ]²
M
.C
Hence,
cosθ ­ sinθ = √2sinθ (option 'D')
=(√3/2).(1/2).(2sin20.sin40).sin80
= (√3/4).(cos20 ­ cos60).sin80
= (√3/8).[2sin10.cos90 + √3/2]
= (√3/8).( 0 + √3/2)
= 3/16 (option 'C')
FO
= (√3/8).[2cos20.sin80 ­ sin80]
= (√3/8).[sin100 ­ sin80 + sin60]
R
U
20. sin20.sin40.sin60.sin80
AM
Another method
sinA.sin2A.sin4A = (1/4).sin3A
Therefore, sin20.sin40.sin60.sin80
EX
= (sin20.sin40.sin80)*sin60
= [(1/4).sin60].sin60
= (1/4)((√3/2)(√3/2)
SS
C
= 3/16 (option 'C')
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