CAT Questions

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Let the distance covered be D km. 

 D10+4+ D10-4 = 5

 D14+D6=5

 10D = 42 x 5 = 210

=> D = 21 km

Ratio of investments for 1 year

=> (P : Q : R) = (2x2 + 2.4x10) : (3x2 + 3.3x10) : (5x12)

=> (P : Q : R) = 28 : 39 : 60

Now R share = 190500 x 60/127 = Rs. 90000.

Succus Entericus is the alkaline secretion produced by glands in the wall of the duodenum, consisting of water, mucoproteins, and hydrogencarbonate ions.

Succus Entericus is secreted by Crypts of Lieberkuhn.

It helps to counteract the highly acidic and proteolytic chyme entering the small intestine from the stomach, and thus protects the duodenum from damage.

Total number of students = 54 x 30
When arranged in rows of 45, number of rows formed are,
54x3045 = 36.

Let the number of students in classes A, B and C be P, Q and R respectively.
Then, total score of A = 83P, total score of B = 76Q, total score of C = 85R.
Also given that,
(83P + 76Q) / (P + Q) = 79
=>4P = 3Q.
(76Q + 85R)/(Q + R) = 81
=>4R = 5Q,
=>Q = 4P/3 and R = 5P/3
Therefore, average score of A, B, C = ( 83P + 76Q + 85R ) / (P + Q + R) = 978/12 = 81.5

False. The ankle is not the third-class lever in the body.

Levers are typically labeled as first class, second class, or third class. All three types are found in the body, but most levers in the human body are third class.

First-class levers in the human body are rare and have the axis (fulcrum) located between the weight (resistance) and the force. One example is the joint between the head and the first vertebra.

In a second-class lever, the weight (resistance) is located between the axis (fulcrum) and the force. An example of a second-class lever in the human body is found in the lower leg when someone stands on tiptoes

In a third-class lever, the most common in the human body, force is applied between the resistance (weight) and the axis (fulcrum). There are numerous third-class levers in the human body; one example can be illustrated in the elbow joint.

Given numbers are 1, 2 and 3

Factors of 1 = 1

Factors of 2 = 1, 2

Factors of 3 = 1, 2 and 3

Common factors = 1

Hence, the HCF of 1, 2 and 3 = 1.