in a certain examination 52 candidate offers biology,60 offers history,96 offers maths, if 21 offered both biology and history,22 offered maths and biology, and 16 offered maths and history, if 7 candidates offered all the subject. how many candidates were there for the examination​

There are 135 candidates.

Step-by-step explanation:

we know that:

52 offer biology

60 offers history

96 offers maths

21 offered both biology and history

22 offered maths and biology

16 offered maths and history

if 7 candidates offered all the subject.

We want to find the total number of candidates.

We start by adding the numbers for each particular exam:

(52 + 60 + 96)

Now, there are 21 that offered both biology and history, then these 21 are counted in the 52 for biology and in the 60 for history, then we need to subtract 21 (because we are counting it two times)

(52 + 60 + 96) - 21

The same happens for the 22 that offered maths and biology, and 16 that offered maths and history, then we get:

(52 + 60 + 96) - 21 - 22 - 16

And there are another 7 that offered for the 3 subjects, then we are counting these 7 ones 3 times, this means that we need to subtract 7 two times.

then we get:

(52 + 60 + 96) - 21 - 22 - 16 - 7 - 7

This is the total number of candidates, if we solve the equation we get:

candidates = (52 + 60 + 96) - 21 - 22 - 16 - 7 - 7 = 135

So we can conclude that there are 135 candidates.

12cm

the answer is d

x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x

step-by-step explanation:

last month maria hiked the 5-mile mountain trail a number of times and she hiked the 10-mile canal trail several times. let x represent the number of times she hiked the 5-mile trail, and let y represent the number of times she hiked the 10-mile trail. if she hiked a total of 90 miles, which equation can be used to find the number of times maria hiked each trail? x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x