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The pie-chart given below shows expenditure incurred by a family on various items and their savings. Study the chart and answer the questions based on the pie-chart
The ratio of expenditure on food to savings is:
Degree of expenditure on food = 120
Degree of expenditure on savings = 60
=> Required ratio = 120/6 = 2 : 1
The graph shows the result of 10th class students of a school for 4 years. Study the graph and answer the questions:
The number of students appeared for the 10th class exam in the year 2002 is
In 2002, no. of students who passed in :
1st class = 120
2nd class = 60
3rd class = 15
=> Total no. of students appeared for class 10th exam in 2002 = 120 + 60 + 15 = 195
The graph shows Income and Expenditure(Rs. in lakhs) of a company.Study the graph and answer the questions.
The expenditure from 2002 to 2003 increased by
Expenditure in 2002 = 30
Expenditure in 2003 = 40
So, the answer is option A.
In the following questions, the following Bar Diagram shows the percentage of 7 coloured balls sold in 5 shops A, B, C, D and E. All the five shops sold an equal number of balls. Study the graph and answer all the questions.
In shop A, the ratio of percentage of violet balls and yellow balls is:
In shop A, percentage of violet balls sold as a percentage of total number of balls = 35%
Percentage of yellow balls sold as a percentage of total number of balls = 20%
Ratio = 35 : 20 = 7 : 4
Option B is the right answer.
Directions: The following graph shows the production of cotton bales of 100 kg each in lakhs by different states A, B, C, D and E over the years. Study the graph and answer the following Questions.
In which State(s) is there a steady increase in the production of cotton during the given period?
Only in A and C there is a steady increment in production of cotton as in D and E, It is decreased and in B production is equal for two years. Hence answer will be C.
Multiplying P inside whole equation will reduced to ((P + 1)3)1/3 = (p + 1)
hence answer will be 125
If x = 2 + √3, y = 2 − √3, then the value of
as x + y = 4
and xy = 1
Now after putting values of x + y and xy and solving, we will get its value as 7/26.
If a2 + b2 + 4c2 = 2(a + b − 2c − 3) and a, b, c are real, then the value of (a2 + b2 + c2) is
a2 + b2 + 4c2 = 2(a + b − 2c − 3)
So, (a − 1)2 + (b − 1)2 + (2c + 1)2 = 0
Or, a = 1, b = 1, c = −1/2
So, a2 + b2 + c2 = 1 + 1 + 1/4 = 9/4
If x2 − 3x + 1, then value of
Given x2 − 3x + 1 = 0
or x2 − 3x = −1
or (x − 3) = −1/x
Now according to given question
After putting value of x + 1/x in above equation, it will get reduced to 10.
The sum of a non-zero number and thrice its reciprocal is 52/7. Find the number.
Let the number be x
According to ques,
=> 7x2 + 21 = 52x
=> 7x2 − 52x + 21 = 0
=> 7x2 − 49x − 3x + 21 = 0
=> 7x(x - 7) - 3(x - 7) = 0
=> (7x - 3)(x - 7) = 0
=> x = 7, 3/7
=> Ans - (C)
The product of two numbers is 45 and their difference is 4. The sum of squares of the two numbers is
As we know (a − b)2 = a2 + b2 − 2ab
We assume that first number is a and second number is b hence ab = 45
and a - b = 4
after putting values we will get a2 + b2 = 106
In Δ ABC, ∠B = 90° and AB : BC = 2 : 1. The value of sin A + cot C is
Given: ∠B = 90 and AB : BC = 2 : 1
To find: sin A + cot C = ?
Solution: Let AB = 2x and BC = x
In right △ABC
Now, sin A + cot C
In a rhombus ABCD, ∠A = 60° and AB = 12 cm. Then the diagonal BD is
Given: ∠A = 60° and AB = 12 cm
To find: BD = ?
Solution: Since, all the sides of a rhombus are equal, => AB = AD
=> ∠ABD = ∠ADB
Now, in △ABD
=> ∠ABD + ∠ADB + ∠A = 180°
=> 2∠ABD = 120°
=> ∠ABD = ∠ADB = ∠A = 60°
=> ABD is equilateral triangle.
=> AB = AD = BD = 12 cm
If there are four lines in a plane, then what cannot be the number of points of intersection of these lines?
Maximum number of intersection points of n lines =
Thus, in a plane of 4 lines, maximum intersection points =
=> 7 is not possible.
=> Ans - (D)
The square root of 14 + 6√5
Given question can be written as 9 + 5 + 6√5
or it will be square of 3 + √5
The difference between the compound interest and simple interest for the amount Rs. 5,000 in 2 years is Rs.32. The rate of interest is
Difference between compound interest and simple interest for 2 years will be
= (where P is principal amount 5000 and r is rate )
after solving above equation we will get r = 8%
In what time will 8,000, at 3% per annum, produce the same interest as 6, 000 does in 5 years at 4 % simple interest?
It is given that Rs 6000 with interest 4% per annum (SI) produces an interest in 5 years =
= Rs 1200
Now the new principal amount is Rs 8000 and rate of interest is 3 % per annum. Let the time be T years in which same Rs 1200 will be generated as interest.
T = 5 years
The compound interest on a certain sum of money for 2 years at 5% per annum is 410. The simple interest on the same sum at the same rate and for the same time is
We know that:
1. For first year, compound interest and simple interest is same if the principal amount and rate of interest is same in both cases.
2. From 2nd year onwards, the compound interest is normal interest plus the interest on accumulated amount due to interest until last cycle.
3. Every year simple interest remains same if Rate of Interest and principal amount remains same.
Let the compound interest for 1st year be Rs y
For two years, CI = Rs 410
y = 200
So for two years , Simple Interest = 200 + 200 = Rs 400
The compound interest on a certain sum of money at a certain rate per annum for two years is 2,050, and the simple interest on the same amount of money at the same rate for 3 years is 3, 000. Then the sum of money is
let the sum of money be Rs P
And rate of interest = R %
It is given that for two years the Compound Interest = Rs 2050
As it is given that for 3 years the Simple Interest = Rs 3000 and we know that SI for every year is same so for 1st year
Simple interest = Rs 1000
For 1st year SI and CI are same if rate of interest and Principal amount is same and hence for 1st year CI = 1000
For 2 years the CI = 2050 = 1000 + 1000 + R/100 x 1000)
R = 5%
P = Rs 20000
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 18% per annum is Rs 81. The sum is
Let the given sum = Rs. 100x
Rate of interest = 18% and time period = 2 years
=> Difference between simple and compound interests =
∴ Value of given sum = 100 × 25 = Rs.2500
The H.C.F. and L.C.M. of, two numbers are 8 and 48 respectively. If one of the numbers is 24, then the other number is
Numbers: First = 24
Second = x (suppose)
H.C.F. of numbers = 8
L.C.M. of numbers = 48
As we know:
H.C.F. * L.C.M. = Product of numbers
48 * 8 = 24 * x
x = 16
L.C.M. of two numbers is 120 and their H.C.F. is 10. Which of the following can be the sum of those two numbers?
We assume that numbers are hr1 and hr2 (where h = H.C.F. of numbers and r1 and r2 are prime factors)
So L.C.M. will be = hr1r2 = 120
or r1r2 = 12
So r1 = 4 and r2 = 3; numbers will be 40 and 30, sum is 70
or r1 = 12 and r2 = 1; numbers will be 120 and 10, sum is 130
Hence only option D justifies.
Two numbers are in the ratio 3 : 4. Their L.C.M. is 84. The greater number is
Let the numbers be 3x, 4x
LCM of 3x and 4x is = 12x
So the number 84 is divisible by 12
84/12 = 7
The numbers are 7x3 = 21 , 7x 4 = 28
The greatest number is 28
What is the HCF (highest common factor) of 57 and 513?
Factors of 57 = 1, 3, 19, 57
Factors of 513 = 1, 3, 9, 19, 27, 57, 171, 513
The common factors are = 1, 3, 19, 57
=> Highest common factor = 57
=> Ans - (B)
The sum of two numbers is 36 and their H.C.F and L.C.M. are 3 and 105 respectively. The sum of the reciprocals of two numbers is
Let's say numbers are x and y
hence sum of the reciprocals will be 1/x + 1/y
as x + y = 36 (given)
and xy = HCF × LCM
= 3 × 105 = 315
after putting the values we will get summation of reciprocals equals to 4/35