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*Answer can only contain numeric values

QUESTION: 1

Find r when C(n,r) = 35 and P(n,r) = 840

Solution:

From the given conditions,

we get

C(n,r) = n!/[r!(n-r)!] = 35...(1)

P(n,r) = n!/(n-r)! = 840...(2)

Dividing (2) by (1), we get n!/(n-r)!/n!/[r!(n-r)!]

= 840/35 r! = 24 r! = 1*2*3*4 = 24

r = 4

*Answer can only contain numeric values

QUESTION: 2

If (n+1)! = 30*(n-1)!, then find n.

Solution:

(n+1)! = 30*(n-1)! (n+1)*n*(n-1)!

= 30(n-1)! (n+1)*n

= 30 n^{2} + n - 30

= 0 n^{2} +6n-5n - 30

=0 n(n+6) - 5(n+6)

= 0 n = -6,

5 Since n cannot be negative,

n = 5

*Answer can only contain numeric values

QUESTION: 3

Find x if (0,0), (3, sqrt(3)) and (x, 2*sqrt(3)) are the vertices of an equilateral triangle in the first quadrant.

Solution:

Let the vertices (0,0), (3, sqrt(3)) and (x, 2*sqrt(3)) be A, B and C respectively.

Since the triangle is equilateral, we have

AB = BC = CA AB^2 = BC^2 = CA^2 (3-0)^2+(sqrt(3)-0)^2

= (3-x)^2 + (2*sqrt(3) - sqrt(3))^2

= (x-0)^2 + (2*sqrt(3)-0)^2 9 + 3

= 9 - 2x + x^2 +3

= x^2 + 12

Hence,

x^2+12 = 12 x = 0 [AB^2=AB*AB]

*Answer can only contain numeric values

QUESTION: 4

C can complete the work in 18 days if he works alone. B takes 3 days lesser than C and A takes 5 days lesser than B to complete the work. On which day will they complete the work if C works for the initial two days only?

Solution:

C takes 18 days to complete the work. B takes 3 days lesser.

Hence, B takes 18-3 = 15 days.

A takes 5 days lesser than B.

Hence, A takes 15-5 = 10 days.

Work done by A in one day = 1/10

Work done by B in one day = 1/15

Work done by C in one day = 1/18

Work done by the three of them in two days

= 2(1/18+1/10+1/15)

= 2(5+9+6)/90

= 2*20/90

= 4/9 Remaining work

= 1-4/9 = 5/9

Work done by A and B in one day

= 1/10+1/15

= (3+2)/30

=1/6

Time taken by A and B to complete the remaining work

= 5/9*6/1

= 10/3 = 3.33

The work in completed on

2+3.33 = 6th day

*Answer can only contain numeric values

QUESTION: 5

If A and B are two mutually exclusive events and P(A) = 2/3 and P(B) = 4/9, then find the probability of the occurrance of A and B together.

Solution:

Since A and B are mutually exclusive events, they do not occur together at all.

Hence, the required probability is zero.

*Answer can only contain numeric values

QUESTION: 6

Find the diagonal of a cuboid whose volume is 144 cc and the dimensions of its base are 12cm and 4cm.

Solution:

The length (l) and breadth (b) of the base are 12 cm and 4 cm respectively.

Height (h) of the cuboid = volume/(l*b) = 144/(12*4) = 3 cm

Diagonal of the cuboid = sqrt(l^{2}+b^{2}+h^{2})

= sqrt (12^{2}+4^{2}+3^{2})

= sqrt (144+16+9)

= sqrt(169)

= 13 cm

The diagonal of the cuboid is 13 cm.

[l^{2}=l*l]

*Answer can only contain numeric values

QUESTION: 7

Find the value of k for which both the equations 12x^{2}+4kx+3=0 and (k+1)x^{2}-2(k-1)x+1=0 have equal real roots. [x^{2}=x*x]

Solution:

Since the two equations have equal real roots, their discriminants are 0.

For 12x^{2}+4kx+3=0, D = 0

D^2 = 0 b^{2}-4ac=0

(4k)^{2} - 4*12*3 = 0 16k^{2} - 144=0

k^{2} = 144/16 = 9

k = -3,

3 For (k+1)x^{2}-2(k-1)x+1=0

b^{2}-4ac=0 [2*(k-1)]^{2}-4*(k+1)*1=0

4(k^{2}-2k+1)-4(k+1)=0

4[k^{2}-2k+1-k-1]=0

k^{2} - 3k = 0 k(k-3)=0 k=0,3

Hence, we have k = 3 [k^{2}=k*k]

*Answer can only contain numeric values

QUESTION: 8

What percent is the first number of the second if the first number is 120% of the third and the second number is 150% of the third?

Solution:

Let the first, second and the third numbers be x, y and z respectively.

x = 120% of z = 120/100*z

y = 150% of z = 150/100*z

x/y*100 = (120z/100)/(150z/100)*100

= 80% x is 80% of y.

*Answer can only contain numeric values

QUESTION: 9

The difference between two numbers is 2. Each number is less than 14 and their sum is greater than 22. Find the greater number of the two.

Solution:

Let the two numbers be x and y such that x < y.

y ??? x = 2,

x < 14, y < 14, x + y > 22 y

= x + 2, x < 14, x + 2 < 14, x + x + 2 > 22 x < 14, x < 12, 2x + 2 > 22 x < 12, x +1 > 11 x < 12, x > 10

Hence, x can be 11 and the corresponding value of y is 13.

The two numbers are 11 and 13.

*Answer can only contain numeric values

QUESTION: 10

The sum of the squares of the two positive numbers is 68 and the square of their difference is 36. Find the sum of the numbers.

Solution:

Let the numbers be x and y,

x>y x^{2}+y^{2}=68 and (x-y)^{2}

=36 (x-y)^{2}

= x^{2}+y^{2}-2xy 36

= 68-2xy 2xy

=68-36=32 xy

=32/2=16 (x+y)^{2}

=x^{2}+y^{2}+2xy = 68+2*16

=68+32=100 x+y

= sqrt(100)

= 10 [x^{2}=x*x]

*Answer can only contain numeric values

QUESTION: 11

The difference between the time when the lightening was seen and the time when the thunder was heard is 10 seconds. Sound covers a distance 330 meters in one second. Find the distance of the thundercloud from the point of observation in meters.

Solution:

Distance = speed*time

= 330*10

= 3300 meters

*Answer can only contain numeric values

QUESTION: 12

Six points lie on a circle. How many cyclic quadrilaterals can be formed by joining the points?

Solution:

Since, the points lie on a circle, all the possible quadrilaterals shall be cyclic.

Number of quadrilateals = C(6,4)

= 6!/(2!4!) = 6*5/2 = 15

Hence,

15 cyclic quadrilaterals can be drawn.

*Answer can only contain numeric values

QUESTION: 13

When a number 'a' is increased by 17, it equals 60 times its reciprocal. How many values of 'a' are possible?

Solution:

According to the conditions,

we have a+17=60*1/a

a^{2}+17a-60=0

a^{2} +20a-3a-60=0

a(a+20)-3(a+20)=0

(a-3)(a+20)=0

a=3,

-20

When a = 3,

a+17 = 3+17 = 20 = 60*1/3

When a = -20,

a+17

= -20+17

= -3 = 60*(-1/20)

Hence, the two values of 'a' are valid.

There are two possible values of 'a'.

[a^{2}=a*a]

*Answer can only contain numeric values

QUESTION: 14

The tax on a commodity decreases by 10% and the consumption increases by 20%. What is the percentage increase in the revenue?

Solution:

Revenue = Tax*Consumption Let the original tax be x and consumption be y.

According to the given conditions, the new tax and consumption will be (x-10x/100) = 90x/100 and (y+20y/100) = 120y/100

The original revenue would be xy and the new revenue would be

(90x/100)*(120y/100)

Perentage change in revenue =

(new revenue-old revenue)/old revenue * 100

= (90x/100*120x/100-xy)/xy*100

= (1.08xy-xy)/xy*100

= 0.08*100

= 8%

The increase in revenue will be 8%.

*Answer can only contain numeric values

QUESTION: 15

A shopkeeper bought flowers at the rate of 8 for Rs.34. He sold them at 12 flowers for Rs.57 and gained Rs. 900. How many flowers were there?

Solution:

Let x be the number of flowers bought and then sold by the shopkeeper.

Let C.P. and S.P. be the cost price and selling price.

Total C.P. = 34/8*x

= 17x/4 Total S.P.

= 57/12*x = 57x/12 Gain

= Total SP - Total CP 900

= 57x/12 - 17x/4 (57x-51x)/12

= x/2 = 900 x = 900*2=1800

Number of flowers = 1800.

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